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92
Chapter 5
Exercise 5B
1 Work out the gradients of these lines:
a y = −2
x + 5 b y = −x
+ 7 c y = 4
+ 3x
d y = 1 _ 3 x − 2 e y = − 2 _ 3 x f y = 5 _ 4 x + 2 _ 3
g 2x −
4y + 5 = 0 h 10x −
5y + 1 = 0 i −x +
2y − 4 = 0
j −3x
+ 6y + 7 = 0 k 4x +
2y − 9 = 0 l 9x +
6y + 2 = 0
2 These lines cut the y-axis a
t (0, c). Work out the value of c in each case.
a y = −x
+ 4 b y = 2
x − 5 c y = 1 _ 2 x − 2 _ 3
d y = −3
x e y = 6 _ 7 x + 7 _ 5 f y = 2 − 7x
g 3x −
4y + 8 = 0 h 4x −
5y − 10 = 0 i −2x
+ y − 9 = 0
j 7x +
4y + 12 = 0 k 7x −
2y + 3 = 0 l −5x
+ 4y + 2 = 0
3 Write these lines in the for
m ax + by + c = 0.
a y = 4
x + 3 b y = 3
x − 2 c y = −6
x + 7
d y = 4 _ 5 x − 6 e y = 5 _ 3 x + 2 f y = 7 _ 3 x
g y = 2
x − 4 _ 7 h y = −3 x + 2 _ 9 i y = −6 x − 2 _ 3
j y = − 1 _ 3 x + 1 _ 2 k y = 2 _ 3 x + 5 _ 6 l y = 3 _ 5 x + 1 _ 2
4 The line y =
6x − 18 meets the x-axis at the point P. Work out the coordinates of P.a 4x − y + 3 = 0
b 1 __ 2 x + y − 5 = 0
x + 2 y − 10 = 0Example 4
Write these lines in the form ax + by + c = 0
a y =
4x + 3 b y =
−1 _ 2 x + 5
Rearrange the equation into the form
ax + by + c = 0
Collect all the terms on one side of the equation.
4x – 8 = 0
4x =
8
x
= 2
So P has coordinates (2, 0)Example 5
The line y = 4 x − 8 meets the x-axis at the point P. Work out the coordinates of P.
The line meets the x-axis when y = 0, so
substitute y = 0 into y = 4x − 8.
Always write down the coordinates of the point.Rearrange the equation for x.
|
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93Straight line graphs
5 The line 3x +
2y = 0 meets the x-axis at the point R. Work out the coordinates of R.
6 The line 5x −
4y + 20 = 0 meets the y-axis at the point A and the x-axis at the point B.
Work out the coordinates of A and B.
7 A line l passes thr
ough the points with coordinates (0, 5) and (6, 7).
a Find the gradient of the line
.
b Find an equation of the line in the f
orm ax + by + c = 0.
8 A line l cuts the
x-axis at (5, 0) and the y-axis at (0, 2).
a Find the gradient of the line
. (1 mark)
b Find an equation of the line in the f
orm ax + by + c = 0. (2 marks)
9 Show that the line with equa
tion ax + by + c = 0
has gradient − a __ b and cuts the y-axis at − c __ b
10 The line l with gr
adient 3 and y-intercept (0, 5)
has the equation ax − 2y + c = 0.
Find the values of a and c. (2 marks)
11 The straight line l
passes through (0, 6) and has gradient −2. It intersects the line with
equation 5x − 8y − 15 = 0 at point P. Find the coordinates of P. (4 marks)
12 The straight line l1 with equation y = 3x − 7 intersects the straight line l2 with equation
ax + 4y − 17 = 0 at the point P(−3, b).
a Find the value of
b. (1 mark)
b Find the value of
a. (2 marks)E
P
E/PTry solving a similar problem with numbers
first:
Find the gradient and y -intercept of the
straight line with equation 3 x + 7 y + 2 = 0.Problem-solving
E/P
E/P
Show that the equation of a straight line through (0, a) and ( b, 0) is ax + by − ab = 0.Challenge
5.2 Equations of straight lines
You can define a straight line by giving:
• one point on the line and the gradient
• two different points on the line
You can find an equation of the line from either of these conditions.
■ The equation of a line with gradient m that passes through
the point with coordinates ( x 1 , y 1 ) can be written as
y − y 1 = m(x − x 1 ) .y
x O(x, y)This is any point
on the line
This is thepoint onthe lineyou know(x
1, y1)x – x 1y – y 1
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] |
94
Chapter 5
Example 7
Find the equation of the line that passes through the points (5, 7) and (3, −1).
m = y2 − y1 _______ x2 − x1 = 7 − (−1) ________ 5 − 3 = 8 __ 2 = 4
S
o y −
y1 = m(x − x1)
y + 1 = 4( x – 3)
y + 1 = 4 x − 12
y = 4 x – 13First find the slope of the line.
Here (x1, y1) = (3, −1) and (x2, y2) = (5, 7).
(x1, y1) and (x2, y2) have been chosen to make the
denominators positive.
You know the gradient and a point on the line, so
use y − y 1 = m(x − x 1 ) .
Use m
= 4, x 1 = 3 and y 1 = −1.
Exercise 5CExample 6
Find the equation of the line with gradient 5 that passes through the point (3, 2).
y − 2 = 5( x − 3)
y − 2 = 5 x − 15
y = 5 x − 13
1 Find the equation of the line with gr adient m that passes through the point (x1, y1) when:
a m = 2 and (
x1, y1) = (2, 5) b m = 3 and ( x1, y1) = (−2, 1)
c m =
−1 and (x1, y1) = (3, −6) d m = −4 and (x1, y1) = (−2, −3)
e m = 1 _ 2 and (x1, y1) = (−4, 10) f m = − 2 _ 3 and (x1, y1) = (−6, −1)
g m = 2 and (
x1, y1) = (a, 2a) h m = − 1 _ 2 and (x1, y1) = (−2b, 3b)
2 Find the equations of the lines tha
t pass through these pairs of points:
a (2, 4) and (3, 8) b (0, 2) and (3, 5)
c (−2, 0) and (2, 8) d (5,
−3) and (7, 5)
e (3, −1) and (7, 3) f (−4,
−1) and (6, 4)
g (−1,
−5) and (−3, 3) h (−4,
−1) and (−3, −9)
i ( 1 _ 3 , 2 _ 5 ) and ( 2 _ 3 , 4 _ 5 ) j (− 3 _ 4 , 1 _ 7 ) and ( 1 _ 4 , 3 _ 7 )
3 Find the equation of the line
l which passes through the points A(7, 2) and B(9, −8).
Give your answer in the form ax + by + c = 0. (3 marks)
4 The vertices of the triangle
ABC have coordinates A(3, 5), B(−2, 0) and C(4, −1).
Find the equations of the sides of the triangle. In each case
fin
d the gradient m
then use
y − y 1 = m(x − x 1 ) . Hint
EThis is in the form y − y1 = m(x − x1). Here m = 5
and (x1, y1) = (3, 2). Explore lines of a given gradient
pa
ssing through a given point using GeoGebra.Online
|
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] |
95Straight line graphs
5 The straight line l
passes through (a, 4) and
(3a, 3). An equation of l is x + 6y + c = 0.
Find the value of a and the value of c. (3 marks)
6 The straight line l
passes through (7a, 5) and (3a, 3).
An equation of l is x + by − 12 = 0.
Find the value of a and the value of b. (3 marks)E/P
E/PIt is often easier to find unknown
values in the order they are given
in the question. Find the value of a first then find the value of c .Problem-solving
Consider the line passing through points ( x1, y1) and ( x2, y2).
a Wri
te down the formula for the gradient, m , of the line.
b Sho
w that the general equation of the line can be written in the form y −
y1 ______ y2 − y1 = x − x1 ______ x2 − x1
c Use the equation from part b to find the equation of the line passing through the
po
ints (−8, 4) and ( −1, 7).Challenge
0 = 3 x − 9 so x = 3. A is the point (3, 0).
y − 0 = 2 __ 3 (x − 3)
3y = 2x − 6
−2x + 3 y + 6 = 0Example 8
The line y = 3x − 9 meets the x -axis at the point A . Find the equation of the line with gradient 2 _ 3 that
passes through point
A. Write your answer in the form ax + by + c = 0, where a , b and c are integers.
Use y − y1 = m(x − x1). Here m = 2 _ 3 and
(x1, y1) = (3, 0).
Rearrange the equation into the form
ax + by + c = 0.The line meets the x-axis when y = 0, so substitute y = 0 into y = 3x − 9.
Example 9
The lines y = 4x − 7 and 2x + 3y − 21 = 0 intersect at the point A . The point B has coordinates (− 2, 8).
Find the equation of the line that passes through the points A and B . Write your answer in the form
ax + by + c = 0, where a , b and c are integers.
2x + 3(4 x − 7) − 21 = 0
2x + 12 x − 21 − 21 = 0
14x = 42
x = 3
y = 4(3) − 7 = 5 so A is the point (3, 5).
m = y2 − y1 _______ x2 − x1 = 8 − 5 _______ −2 − 3 = 3 ___ −5 = − 3 __ 5
y − 5 = − 3 __ 5 (x − 3)
5y −
25 = − 3x + 9
3x + 5 y − 34 = 0Solve the equations simultaneously to find point A .
Substitute y = 4x − 7 into 2x
+ 3y − 21 = 0.
Find the slope of the line connecting A and B.
Use y − y 1 = m(x − x 1 ) with m = − 3 __ 5 and
( x 1 , y 1 ) = (3, 5) . Check solutions to simultaneous
eq
uations using your calculator.Online Plot the solution on a graph using
GeoGe
bra.Online
|
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] |
96
Chapter 5
1 The line y =
4x − 8 meets the x-axis at the point A. Find the equation of the line with gradient
3 that passes through the point A.
2 The line y = −2 x + 8 meets the
y-axis at the point B . Find the equation of the line with gradient
2 that passes through the point B.
3 The line y =
1 _ 2 x + 6 meets the x-axis at the point C. Find the equation of the line with gradient 2 _ 3
that passes through the point
C. Write your answer in the form ax + by + c = 0, where a, b and
c are integers.
4 The line y =
1 _ 4 x + 2 meets the y-axis at the point B. The point C has coordinates (−5, 3).
Find the gradient of the line joining the points B and C.
5 The line that passes through the points (2,
−5) and (−7, 4)
meets the x-axis at the point P. Work out the coordinates
of the point P.
6 The line that passes through the points (
−3, −5) and (4, 9) meets the y-axis at the point G.
Work out the coordinates of the point G.
7 The line that passes through the points (3, 2 1 _ 2 ) and (−1 1 _ 2 , 4) meets the y-axis a t the point J.
Work out the coordinates of the point J.
8 The lines y =
x and y = 2x − 5 intersect at the point A. Find the equation of the line with
gradient 2 _ 5 that passes through the point A.
9 The lines y =
4x − 10 and y = x − 1 intersect at the point T. Find the equation of the line with
gradient − 2 _ 3 that passes through the point T. Write your answer in the form ax + by + c = 0,
wher
e a, b and c are integers.
10 The line p has gr
adient 2 _ 3 and passes through the point (6, − 12). The line q has gradient −1 and
passes through the point (5, 5). The line p meets the y-axis at A and the line q meets the x-axis
at B. Work out the gradient of the line joining the points A and B.
11 The line y = −
2x + 6 meets the x-axis at the point P. The line y = 3 _ 2 x − 4 meets the y-axis at the
point Q. Find the equation of the line joining the points P and Q.
12 The line y =
3x − 5 meets the x-axis at the point M. The line y = − 2 _ 3 x + 2 _ 3 meets the y-axis a t the
point N. Find the equation of the line joining the points M and N. Write your answer in the
form ax + by + c = 0, where a, b and c are integers.
13 The line y = 2
x − 10 meets the x-axis at the point A. The line y = −2x + 4 meets the y-axis at the
point B . Find the equa
tion of the line joining the points A and B.
14 The line y =
4x + 5 meets the y-axis at the point C. The line y = −3x − 15 meets the x-axis at
the point D. Find the equation of the line joining the points C and D. Write your answer in the form ax + by + c = 0, where a, b and c are integers.
15
The lines y =
x − 5 and y = 3x − 13 intersect at the point S. The point T has coordinates (−4, 2).
Find the equation of the line that passes through the points S and T.
16 The lines y = −2 x + 1 and
y = x + 7 intersect at the point L . The point M has coordinates (− 3, 1).
Find the equation of the line that passes through the points L and M.P
P
P
P
P
P
P
P
P
P
P
P
PExercise 5D
A sketch can help you check whether
your answer looks right.Problem-solving
|
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] |
97Straight line graphs
Example 10
A line is parallel to the line 6x + 3y − 2 = 0 and it passes through the point (0, 3).
Work out the equation of the line.
6x + 3 y − 2 = 0
3y − 2 = − 6x
3y = −6x + 2
y = −2x + 2 __ 3
The g
radient of this line is − 2.
The equation of the line is y = − 2x + 3.Rearrange the equation into the form y = mx + c
to find m.
Compare y = −2x + 2 _ 3 with y = mx + c, so m = −2.
Parallel lines have the same gradient, so the
gradient of the required line = −2.
(0, 3) is the intercept on the y-axis, so c = 3.
1 Work out whether each pair of lines is parallel.
a y =
5x − 2 b 7x + 14y − 1 = 0 c 4x − 3y − 8 = 0
15x − 3y + 9 = 0 y = 1 _ 2 x + 9 3x − 4y − 8 = 0
2 The line r passes thr
ough the points (1, 4) and (6, 8) and the line s passes through the points
(5, −3) and (20, 9). Show that the lines r and s are parallel.
3 The coordinates of
a quadrilateral ABCD are A(−6, 2), B(4, 8),
C(6, 1) and D(−9, −8). Show that the quadrilateral is a trapezium.
4 A line is paralle
l to the line y = 5x + 8 and its y-intercept is (0, 3).
Write down the equation of the line.
5 A line is paralle
l to the line y = − 2 _ 5 x + 1 and its y-intercept is (0, −4). Work out the equation of
the line. Write your answer in the form ax + by + c = 0, where a, b and c are integers.
6 A line is paralle
l to the line 3x + 6y + 11 = 0 and its intercept on the y-axis is (0, 7). Write down
the equation of the line.
7 A line is paralle
l to the line 2x − 3y − 1 = 0 and it passes through the point (0, 0). Write down the
equation of the line.
8 Find an equation of the line tha
t passes through the point (−2, 7) and is parallel to the line
y = 4x + 1. Write your answer in the form ax + by + c = 0.P
A trapezium has exactly
on
e pair of parallel sides.HintP
The line will have
gra
dient 5.Hint
P
PExercise 5E5.3 Parallel and perpendicular lines
■ Parallel lines have the same gradient.y
x O
|
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] |
98
Chapter 5
Example 11
Work out whether these pairs of lines are parallel, perpendicular or neither:
a 3x − y − 2 = 0 b y = 1 _ 2 x
x + 3y − 6 = 0 2x − y + 4 = 0
a 3x − y − 2 = 0
3x − 2 = y
So y =
3x − 2
The gradient of this line is 3.
x + 3 y − 6 = 0
3y − 6 = − x
3y = − x + 6
y = − 1 __ 3 x + 2
The gradient of this line is − 1 __ 3
So th
e lines are perpendicular as
3 × (− 1 __ 3 ) = −1.
b y = 1 __ 2 x
The g
radient of this line is 1 __ 2
2x
− y + 4 = 0
2x + 4 = y
So y =
2x + 4
The gradient of this line is 2.
The lines are not parallel as they have
different gradients.The lines are not perpendicular as
1 __ 2 × 2 ≠ −1.Rearrange the equations into the form y = mx + c.
Compare y = − 1 __ 3 x + 2 with y = mx + c, so m = − 1 _ 3 Perpendicular lines are at right angles to each other.
If you know the gradient of one line, you can find thegradient of the other.
■
If a line has a gradient o
f m, a line
perpendicular to it has a gradient of − 1 __ m
■ If two lines are perpendicular, the product
of their gr
adients is −1.l1
l2–b
ba
aThe shaded triangles
are congruent.
Line l 1 has gradient
a __ b = m
Line l 2 has gradient
−b ___ a = − 1 __ m
Compare y = 1 _ 2 x with y = mx + c, so m = 1 _ 2 .
Rearrange the equation into the form y = mx + c
to find m.
Compare y = 2x + 4 with y = mx + c, so m = 2.
Explore this solution using
GeoGe
bra.Online
|
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99Straight line graphs
Example 12
A line is perpendicular to the line 2y − x − 8 = 0 and passes through the point (5, −7).
Find the equation of the line.
y = 1 __ 2 x + 4
Grad
ient of y = 1 __ 2 x + 4 is 1 __ 2
So th
e gradient of the perpendicular line is − 2.
y − y 1 = m(x − x 1 )
y +
7 = − 2(x − 5)
y + 7 = − 2x + 10
y = −2x + 3You need to fill in the steps of this problem
yourself:
• Rearrange the equation into the form y =
mx + c to find the gradient.
• Use − 1 __ m to find the gradient of a perpendicular
line.
• Use y − y1 = m(x − x1) to find the equation of
the line.Problem-solving
1 Work out w
hether these pairs of lines are parallel, perpendicular or neither:
a y =
4x + 2 b y = 2 _ 3 x − 1 c y = 1 _ 5 x + 9
y = − 1 _ 4 x − 7 y = 2 _ 3 x − 11 y = 5 x + 9
d y =
−3x + 2 e y = 3 _ 5 x + 4 f y = 5 _ 7 x
y = 1 _ 3 x − 7 y = − 5 _ 3 x − 1 y = 5 _ 7 x − 3
g y =
5x − 3 h 5x − y − 1 = 0 i y = − 3 _ 2 x + 8
5x − y + 4 = 0 y = − 1 _ 5 x 2x − 3y − 9 = 0
j 4x
− 5y + 1 = 0 k 3x + 2y −12 = 0 l 5x − y + 2 = 0
8x − 10y − 2 = 0 2x + 3y − 6 = 0 2x + 10y − 4 = 0
2 A line is perpendicular to the line y
= 6x − 9 and passes through the point (0, 1). Find an
equation of the line.
3 A line is perpendicular to the line 3x
+ 8y − 11 = 0 and passes through the point (0, −8).
Find an equation of the line.
4 Find an equation of the line tha
t passes through the point (6, −2) and is perpendicular to the
line y = 3x + 5.
5 Find an equation of the line tha
t passes through the point (−2, 5) and is perpendicular to the
line y = 3x + 6.
6 Find an equation of the line tha
t passes through the point (3, 4) and is perpendicular to the line
4x − 6y + 7 = 0.
7 Find an equation of the line tha
t passes through the point (5, −5) and is perpendicular to the
line y = 2 _ 3 x + 5. W rite your answer in the form ax + by + c = 0, where a, b and c are integers.P
PExercise 5F
|
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] |
100
Chapter 5
8 Find an equation of the line tha
t passes through the point (−2, −3) and is perpendicular to the
line y = − 4 _ 7 x + 5. W rite your answer in the form ax + by + c = 0, where a, b and c are integers.
9 The line l passes thr
ough the points (−3, 0) and (3, −2)
and the line n passes through the points (1, 8) and (−1, 2).
Show that the lines l and n are perpendicular.
10 The vertices of a quadrila
teral ABCD have coordinates
A(−1, 5), B(7, 1), C(5, −3) and D(−3, 1). Show that the quadrilateral is a rectangle.
11
A line l 1 has equation 5x + 11y − 7 = 0 and crosses the x-axis at A. The line l 2 is perpendicular
to l 1 and passes through A.
a Find the coordinates of
the point A. (1 mark)
b Find the equation of the line l 2 . Write your answer in the form ax + by + c = 0. (3 marks)
12 The points A and
C lie on the y-axis and the point B lies on the x-axis as shown in the diagram.
The line through points A and B is perpendicular to the line through points B and C.
Find the value of c. (6 marks)P
The sides of a rectangle
ar
e perpendicular.HintP
E/P
E/P
Sketch graphs in coordinate geometry problems
are not accurate, but you can use the graph to make sure that your answer makes sense. In this question c must be negative. Problem-solving
Oy
xA (0, 4)
C (0, c)B (–3, 0)Don’t do more work than you need to. You only need to find the gradients of both lines, not their equations.Problem-solving
5.4 Length and area
You can find the distance between two points A and B by
considering a right-angled triangle with hypotenuse AB.
■ You can find the distance d between ( x 1 , y 1 ) and ( x 2 , y 2 )
by using the f
ormula
d = √ _________________ ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 y
x O(x2, y2)
(x1, y1)x2 – x 1y2 – y 1 d
AB
|
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] |
101Straight line graphs
Example 13
Find the distance between (2, 3) and (5, 7).
y
x Od
(2, 3)/four.ss01
3(5, 7)
d 2 = (5 − 2)2 + (7 − 3)2
d 2 = 32 + 42
d = √ __________ (32 + 42)
= √ ___ 25
= 5Draw a sketch.
Let the distance between the points be d.
The difference in the y-coordinates is 7 − 3 = 4.
The difference in the x-coordinates is 5 − 2 = 3.
d = √ ___________________ (x2 − x1)2 + (y2 − y1)2 with
(x1, y1) = (2, 3) and ( x2, y2) = (5, 7).
The straight line l 1 with equation 4x − y = 0 and
the straight line l 2 with equation 2x + 3y − 21 = 0 intersect at point A.
a Work out the coor
dinates of A.
b Work out the ar
ea of triangle AOB where B is the point where l 2 meets the x-axis.Example 14
a Equation of l 1 is y = 4 x
2x + 3 y − 21 = 0
2x + 3(4 x) − 21 = 0
14x − 21 = 0
14x = 21
x = 3 __ 2
y = 4 × ( 3 __ 2 ) = 6
So po
int A has coordinates ( 3 __ 2 , 6) .
b The t
riangle AOB has a height of 6 units.
2x + 3 y − 21 = 0
3x + 3(0) − 21 = 0
2x − 21 = 0
x = 21 ___ 2
The t
riangle AOB has a base length of 21 ___ 2
units.
Area = 1 __ 2 × 6 × 21 ___ 2 = 63 ___ 2 The height is the y -coordinate of point A.
B is the point where the line l 2 intersects the x -axis.
At B , the y -coordinate is zero.Substitute to find the y -coordinate of point A.Solve the equation to find the x -coordinate of
point A .Substitute y = 4 x into the equation for l 2 to find
the point of intersection.
Solve the equation to find the x -coordinate of point B .
Area = 1 _ 2 × ba se × height
You don’t need to give units for length and area
problems on coordinate grids.Rewrite the equation of l 1 in the form y = mx + c. Draw both lines and the triangle
AO
B on a graph using GeoGebra.Online
|
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] |
102
Chapter 5
1 Find the distance between these pairs of points:
a (0, 1), (6, 9) b (4, −6), (9, 6) c (3, 1), (−1, 4)
d (3, 5), (4, 7) e (0, −4), (5, 5) f (−
2, −7), (5, 1)
2 Consider the points A(
−3, 5), B(−2, −2) and C (3, −7).
Deter
mine whether the line joining the points A and B is
congruent to the line joining the points B and C.
3 Consider the points P(11,
−8), Q(4, −3) and R(7, 5). Show that the line segment joining the
points P and Q is not congruent to the line joining the points Q and R.
4 The distance between the points (−
1, 13) and (x, 9) is √ ___ 65 .
Find two possib
le values of x.
5 The distance between the points (2, y
) and (5, 7) is 3 √ ___ 10 . Find two possible values of y.
6 a Show that the str
aight line l 1 with equation y = 2x + 4 is
parallel to the straight line l 2 with equation 6x − 3y − 9 = 0.
b Find the equation of the str
aight line l 3 that is perpendicular
to l 1 and passes through the point (3, 10).
c Find the point of intersection of the lines l 2 and l 3 .
d Find the shortest distance between lines l 1 and l 2 .
7 A point P lies on the line with equa
tion y = 4 − 3x. The point P is a distance √ ___ 34 from the
origin. Find the two possib
le positions of point P. (5 marks)
8 The vertices of a triangle ar
e A(2, 7), B(5, −6) and C(8, −6).
a Show that the triangle is a sca
lene triangle.
b Find the area of the triangle
ABC.
9 The straight line l 1 has equation y = 7x − 3. The straight line l 2
has equation 4x + 3y − 41 = 0. The lines intersect at the point A.
a Work out the coor
dinates of A.
The straight line l 2 crosses the x-axis at the point B.
b Work out the coor
dinates of B.
c Work out the ar
ea of triangle AOB . Two line segments are congruent
if th
ey are the same length.Hint
P
P
P
E/P
P Sca lene triangles have
three sides of different lengths.NotationUse the distance formula to
formulate a quadratic equation in x.Problem-solvingExercise 5G
The
s
hortest distance between
two parallel lines is the
perpendicular distance between them.Problem-solving
Draw a sketch and label the points A, B and C.
Find the length of the base and the height of the triangle.Problem-solving
|
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] |
103Straight line graphs
10 The straight line l 1 has equation 4x − 5y − 10 = 0 and intersects the x-axis at point A.
The straight line l 2 has equation 4x − 2y + 20 = 0 and intersects the x-axis at the point B.
a Work out the coor
dinates of A.
b Work out the coor
dinates of B.
The straight lines l 1 and l 2 intersect at the point C.
c Work out the coor
dinates of C.
d Work out the ar
ea of triangle ABC.
11 The points R(5,
−2) and S(9, 0) lie on the straight line l 1 as shown.
a Work out an equa
tion for straight line l 1 . (2 marks)
The straight line l 2 is perpendicular to l 1
and passes through the point R.
b Work out an equa
tion for straight line l 2 . (2 marks)
c Write down the coor
dinates of T. (1 mark)
d Work out the lengths of
RS and TR
leaving your answer in the form k √ __
5 . (2 marks)
e Work out the ar
ea of △RST. (2 marks)
12 The straight line l 1 passes through the point (−4, 14) and has gradient − 1 _ 4
a Find an equation for l 1 in the form ax + by + c = 0, where a, b and c are integers. (3 marks)
b Write down the coor
dinates of A, the point where straight line l 1 crosses the y-axis. (1 mark)
The straight line l 2 passes through the origin and has gradient 3. The lines l 1 and l 2
intersect at the point B.
c Calculate the coor
dinates of B. (2 marks)
d Calculate the e
xact area of △OAB . (2 marks)E
Oy
x
l2l1T
RS
E/P
5.5 Modelling with straight lines
■ Two quantities are in direct proportion when they increase at the same rate. The graph of
these quantities is a straight line thr
ough the origin.
y
x Oy = kx
k
1 The se mean the same thing:
y is proportional to x
y ∝ x
y = kx for some real constant k.Notation
If x increases by
1 unit, y increases
by k units
|
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] |
104
Chapter 5
Example 15
The graph shows the extension, E, of a spring
when different masses, m, are attached to the end of the spring.
a
Calculate the gr
adient, k, of the line.
b Write an equation linking
E and m.
c Explain what the v
alue of k represents in this
situation.E
m100 0 200
Mass on spring (grams)Extension of spring (cm)
300 4005
010152025
You can sometimes use a linear model to show the relationship between two variables, x and y.
The graph of a linear model is a straight line, and the variables are related by an equation of the form y = ax + b.
A linear model can still be appropriate even if all the points do not lie directly on the line. In this case,
the points should be close to the line. The further the points are from the line, the less appropriate a
linear model is for the data.
■
A mathematical model is an attempt to r
epresent a real-life situation using mathematical
concepts. It is often necessary to make assumptions about the real-life problem in order to
create a model.
Example 16
A container was filled with water. A hole was made in the bottom of the container. The depth of water remaining was recorded at certain time intervals. The table shows the results.
Time, t seconds 0 10 30 60 100 120
Depth of water, d cm 19.1 17.8 15.2 11.3 6.1 3.5
a Determine whether a linear model is appropriate by drawing a graph.
b Deduce an equation in the for
m d = at + b.
c Interpret the meaning of the coefficients
a and b.
d Use the model to find the time when the container will be empty
. a slope = 20 − 0 _______ 400 − 0
= 20 ____ 400 = 1 ___ 20
So k
= 1 ___ 20
b E =
km
E = 1 ___ 20 m
c k re
presents the increase in extension in
cms when the mass increases by 1 gram.Use any two points on the line to calculate the
gradient. Here (0, 0) and (400, 20) are used.
‘y = kx ’ is the general form of a direct proportion
equation. Here the variables are E and m .
k is the gradient. When the m -value increases
by 1, the E -value increases by k .Simplify the answer.
|
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] |
105Straight line graphs
a
20 0/four.ss010
Time (seconds)Depth of w/a.ss01ter
d
tDepth of w/a.ss01ter (cm)
60801001202
0/four.ss016810121/four.ss01161820
The points form a straight line, therefore a
linear model is appropriate.
b m = 6.1 − 19.1 ___________ 100 − 0
= − 13 ____ 100 = − 0.13
Th
e d-intercept is 19.1. So b = 19.1
d = at + b
d = −0.13t + 19.1
c a is t
he change in depth of water in the
container every second.b is the depth of water in the container at the beginning of the experiment.
d
d =
−0.13t + 19.1
0 = − 0.13t + 19.1
0.13t = 19.1
t = 146.9 seconds.You need to give your answer in the context of the
question. Make sure you refer to the extension in the spring and the mass.Problem-solving
State the linear equation using the variables in the question.
Substitute d = 0, as we want to know the time
when the depth of water is zero.
Solve the equation to find t .a represents the rate of change. Look at the
problem and determine what is changing every second.
b is the value of d when t = 0. It represents the
starting, or initial, value in the model. Pick any two points from the table. Here (0,
19.1) and (100, 6.1) are used.
Th
e d-intercept is the d- value when t = 0.
State the linear equation using the variables in the question.
Substitute a = − 0.13 and b = 19.1.
Example 17
In 1991 there were 18 500 people living in Bradley Stok e. Planners projected that the number of
people living in Bradley Stoke would increase by 350 each year.
a Write a linear model f
or the population p of Bradley Stoke t years after 1991.
b Write down one r
eason why this might not be a realistic model.
|
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] |
106
Chapter 5
a 1991 is the first year, so t = 0.
Wh
en t = 0, the population is 18 500
.
18 500 i
s the p -intercept.
The population is expected to increase
by 350 each year.
350 represents the gradient of the line.p = at + b
p = 350 t + 18
50
0
b The n
umber of people living in Bradley
Stoke would probably not increase by
exactly the same amount each year.The p -intercept is the population when t = 0.
The gradient is the yearly change in population.
State the linear equation using the variables in
the question.
Substitute a = 350 and b = 18 50 0.
Look at the question carefully. Which points did you accept without knowing them to be true? These are your assumptions .Problem-solving
Exercise 5H
1 For each graph
i calculate the gr
adient, k, of the line
ii write a direct proportion equa
tion connecting the two variables.
a
2 04
Time, t (s)Distance, d (m)
6 810 120100200300400500600 b
10 020
Time skating, t (mins)Cost of skating, C (£)
3002468 c
5 01015 2025
Time reading, t (mins)Pages read, p
300481216
2 Draw a graph to determine whether a linear model would be appropriate for each set of data.
av pbx ycw l
0 0 0 70 3.1 45
15 2 5 82.5 3.4 47
25 6 10 95 3.6 50
40 12 15 107.5 3.9 51
60 25 25 132.5 4.5 51
80 50 40 170 4.7 53
3 The cost of electricity , E, in pounds and the number of kilowatt hours, h, are shown in the table.
kilowatt hours, h 0 15 40 60 80 110
cost of electricity, E 45 46.8 49.8 52.2 54.6 58.2 A linear model can be
ap
propriate even if all the
points do not lie exactly in a
straight line. In these cases, the points should lie close to a straight line.Hint
E/P
|
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] |
107Straight line graphs
a Draw a graph of the data. (3 marks)
b Explain how you kno
w a linear model would be appropriate. (1 mark)
c Deduce an equation in the for
m E = ah + b. (2 marks)
d Interpret the meaning of the coefficients
a and b. (2 marks)
e Use the model to find the cost of 65 kilo watt hours. (1 mark)
4 A racing car acceler
ates from rest to 90 m/s in 10 seconds. The tab
le shows the total distance
travelled by the racing car in each of the first 10 seconds.
time, t seconds 0 1 2 3 4 5 6 7 8 9 10
distance, d m 0 4.5 18 40.5 72 112.5 162 220.5 288 364.5 450
a Draw a graph of the data.
b Explain how you kno
w a linear model would not be appropriate.
5 A website designer char
ges a flat fee and then a daily rate in order to design new websites
for companies.
Company A ’s new website takes 6 days and they are charged £7100.Company B’s new website take 13 days and they are charged £9550.
a
Write an equation linking da
ys, d and website cost, C in the
form C = ad + b. (3 marks)
b Interpret the va
lues of a and b. (2 marks)
c The web designer char
ges a third company £13 400. Calculate the n
umber of days the designer
spent working on the website. (1 mark)
6 The aver
age August temperature in Exeter is 20 °C or 68 °F . The av
erage January temperature in
the same place is 9 °C or 48.2 °F .
a Write an equation linking F
ahrenheit F and Celsius C in the form F = aC + b. (3 marks)
b Interpret the va
lues of a and b. (2 marks)
c The highest temperatur
e recorded in the UK was 101.3 °F . Calcula
te this temperature in Celsius.
(1 mark)
d For wha
t value is the temperature in Fahrenheit the same as the temperature in Celsius?
(3 marks)
7 In 2004, in a city, there w
ere 17 500 homes with internet connections. A service pro
vider predicts
that each year an additional 750 homes will get internet connections.a
Write a linear model f
or the number of homes n with internet connections t years after 2004.
b Write down one assumption made b
y this model.
8 The scatter gra
ph shows the height h and foot length f
of 8 students. A line of best fit is drawn on the scatter
graph.
a Explain why the da
ta can be approximated to a
linear model. (1 mark)
b Use points A and
B on the scatter graph to write a
linear equation in the form h = af + b. (3 mark)
c Calculate the e
xpected height of a person with a
foot length of 26.5 cm. (1 mark) P
E/P
Let ( d 1 , C 1 ) = (6, 7100)
and ( d 2 , C 2 ) = (13, 9550).Hint
E/P
P
222324252627282930 21 20
Foot length, f (cm)Height and foot length
A (24, 165)Height, h (cm)
160165170175180185190
B (27, 177)E/P
|
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] |
108
Chapter 5
1 The straight line passing through the point
P(2, 1) and the point Q (k, 11) has gradient − 5 __ 12
a Find the equation of the line in ter
ms of x and y only. (2 marks)
b Determine the value of k. (2 marks)
2 The points A and
B have coordinates (k, 1) and (8, 2k − 1) respectively, where k is a constant.
Given that the gradient of AB is 1 _ 3
a show that
k = 2 (2 marks)
b find an equation f
or the line through A and B. (3 marks)
3 The line L1 has gradient 1 _ 7 and passes through the point A (2, 2). The line L2 has gradient −1 and
passes through the point B(4, 8). The lines L1 and L2 intersect at the point C.
a Find an equation for
L1 and an equation for L2. (4 marks)
b Determine the coordina
tes of C. (2 marks)
4 a Find an equation of
the line l which passes through the points A(1, 0) and B(5, 6). (2 marks)
The line m with equa
tion 2x + 3y = 15 meets l at the point C.
b Determine the coordina
tes of C. (2 marks)
5 The line L passes thr
ough the points A(1, 3) and B(−19, −19).
Find an equation of L in the form ax + by + c = 0. where a, b and c are integers. (3 marks)
6 The straight line l1 passes through the points A and B with coordinats (2, 2) and (6, 0) respectively.
a Find an equation of
l1. (3 marks)
The straight line l2 passes through the point C with coordinate (−9, 0) and has gradient 1 _ 4 .
b Find an equation of
l2. (2 marks)
7 The straight line l
passes through A(1, 3 √ __
3 ) and B(2 + √ __
3 , 3 + 4 √ __
3 ).
Show that l meets the x-axis at the point C(−2, 0). (5 marks)
8 The points A and
B have coordinates (−4, 6) and (2, 8) respectively. A line p is drawn through B
perpendicular to AB to meet the y-axis at the point C.
a Find an equation of the line
p. (3 marks)
b Determine the coordina
tes of C. (1 mark)E/P
E/P
E
E
E
E
E/P
EMixed exercise 59 The price P of a good and the quantity Q of a good are linked.
The demand for a new pair of trainers can be modelled using the equation P = − 3 _ 4 Q + 35 .
The supply of the tr
ainers can be modelled using the equation P = 2 _ 3 Q + 1 .
a Draw a sk
etch showing the demand and supply lines on the same pair of axes.
The equilibrium point is the point where the supply and demand lines meet.
b Find the values of
P and Q at the equilibrium point.
|
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] |
109Straight line graphs
9 The line l has equa tion 2x − y − 1 = 0.
The line m passes through the point A(0, 4) and is perpendicular to the line l.
a Find an equation of
m. (2 marks)
b Show that the lines
l and m intersect at the point P(2, 3). (2 marks)
The line n passes thr
ough the point B(3, 0) and is parallel to the line m.
c Find the coordinates of
the point of intersection of the lines l and n. (3 marks)
10 The line l1 passes through the points A and B with coordinates (0, −2) and (6, 7) respectively.
The line l2 has equation x + y = 8 and cuts the y-axis at the point C.
The line l1 and l2 intersect at D.
Find the area of triangle ACD . (6 marks)
11 The points A and
B have coordinates (2, 16) and (12, −4) respectively.
A straight line l1 passes through A and B.
a Find an equation for
l1 in the form ax + by = c. (2 marks)
The line l2 passes through the point C with coordinates (−1, 1) and has gradient 1 _ 3
b Find an equation for
l2. (2 marks)
12 The points A(
−1, −2), B(7, 2) and C(k, 4), where k is a constant, are the vertices of △ABC.
Angle ABC is a right angle.
a Find the gradient of
AB. (1 mark)
b Calculate the v
alue of k. (2 marks)
c Find an equation of the str
aight line passing through B and C. Give your answer in
the form ax + by + c = 0, where a, b and c are integers (2 marks)
d Calculate the ar
ea of △ABC. (2 marks)
13 a Find an equation of the str
aight line passing through the points with coordinates
(−1, 5) and (4, −2), giving your answer in the form ax + by + c = 0,
where a, b and c are integers. (3 marks)
The line crosses the x
-axis at the point A and the y-axis at the point B, and O is the origin.
b Find the area of
△AOB . (3 marks)
14 The straight line l1 has equation 4y + x = 0.
The straight line l2 has equation y = 2x − 3.
a On the same axes, sk etch the graphs of l1 and l2. Show clearly the coordinates of all
points at which the graphs meet the coordinate axes. (2 marks)
The lines l1 and l2 intersect at the point A.
b Calculate
, as exact fractions, the coordinates of A. (2 marks)
c Find an equation of the line thr
ough A which is perpendicular to l1.
Give your answer in the form ax + by + c = 0, where a, b and c are integers. (2 marks)E/P
E/P
E
E/P
E/P
E
|
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] |
110
Chapter 5
15 The points A and
B have coordinates (4, 6) and (12, 2) respectively.
The straight line l1 passes through A and B.
a Find an equation for
l1 in the form ax + by + c = 0, where a, b and c are integers. (3 marks)
The straight line l2 passes through the origin and has gradient − 2 _ 3
b Write down an equa
tion for l2. (1 mark)
The lines l1 and l2 intersect at the point C.
c Find the coordinates of
C. (2 marks)
d Show that the lines
OA and OC are perpendicular, where O is the origin. (2 marks)
e Work out the lengths of
OA and OC . Write your answers in the form k √ ___ 13 . (2 marks)
f Hence calculate the ar
ea of △OAC . (2 marks)
16 a Use the distance form
ula to find the distance between (4a, a) and (−3a, 2a).
Hence find the distance between the following pairs of points:
b (4, 1) and (−3, 2) c (12, 3) and (−9, 6) d (−
20, −5) and (15, −10)
17 A is the point (−
1, 5). Let (x, y) be any point on the line y = 3x.
a Write an equation in ter
ms of x for the distance between (x, y) and A(−1, 5). (3 marks)
b Find the coordinates of
the two points, B and C, on the line y = 3x which are a distance
of √ ___ 74 from (−1, 5). (3 marks)
c Find the equation of the line l 1 that is perpendicular to y = 3x and goes through the
point (−1, 5). (2 marks)
d Find the coordinates of
the point of intersection between l 1 and y = 3x. (2 marks)
e Find the area of triangle
ABC. (2 marks)
18 The scatter gra
ph shows the oil production P and carbon dioxide emissions C for various years
since 1970. A line of best fit has been added to the scatter graph.
Oil production (million tonnes)Oil production and carbon dioxide emissionsCarbon dioxide emissions
(million tonnes)
500 1000 1500 2000 2500 3000 3500 4000 4500 00500010 00015 00020 00025 00030 00035 000
a Use two points on the line to calculate its gradient. (1 mark)
b Form
ulate a linear model linking oil production P and carbon dioxide emissions C,
giving the relationship in the form C = aP + b. (2 marks)
c Interpret the va
lue of a in your model. (1 mark)
d With refer
ence to your value of b, comment on the validity of the model for small
values of P. (1 mark)E
E/P
E/P
|
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] |
111Straight line graphs
1 Find the area of the triangle with vertices A(−2, −2), B(13, 8) and C(−4, 14).
2 A tria
ngle has vertices A(3, 8), B(9, 9) and C(5, 2) as shown in
the diagram.
The l
ine l 1 is perpendicular to AB and passes through C.
The l
ine l 2 is perpendicular to BC and passes through A.
The l
ine l 3 is perpendicular to AC and passes through B.
Sho
w that the lines l 1 , l 2 and l 3 meet at a point and find the
coordinates of that point.
3 A tria
ngle has vertices A(0, 0), B(a, b) and C(c, 0) as shown in
the diagram.
The l
ine l 1 is perpendicular to AB and passes through C.
The l
ine l 2 is perpendicular to BC and passes through A.
The l
ine l 3 is perpendicular to AC and passes through B.
Fin
d the coordinates of the point of intersection of l 1 , l 2 and l 3 .Challenge
xy
Ol1
l2l3B (9, 9)
A (3, 8)
C (5, 2)
O xy
l1l2
l3B (a, b)
A (0, 0) C (c, 0)
|
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] |
112
Chapter 5
1 The gradient m of the line joining the point with
coor
dinates (x1, y1) to the point with coordinates
(x2, y2) can be calculated using the formula
m = y2 – y1 ______ x2 – x1
2 ● The equation of a str
aight line can be written in the form
y = mx + c,
where m is the gradient and (0, c) is the y-intercept.
● The equation of a str
aight line can also be written in
the f
orm
ax + by + c = 0,
where a, b and c are integers.
3 The equation of a line with gradient
m that passes through the point with coordinates
( x 1 , y 1 ) can be written as y − y 1 = m(x − x 1 ) .
4 Parall
el lines have the same gradient.
5 If a line has a gradient
m, a line perpendicular to it has a gradient of − 1 __ m
6 If two lines ar
e perpendicular, the product of their gradients is −1.
7 You can find the distance
d between (x1, y1) and (x2, y2) by using the formula
d = √ ___________________ (x2 − x1)2 + (y2 − y1)2 .
8 The point of intersection o
f two lines can be found using simultaneous equations.
9 Two quantities ar
e in direct proportion when they increase at the same rate.
The graph of these quantities is a straight line through the origin.
10 A mathematical model is an attempt to r
epresent a real-life situation using mathematical
concepts. It is often necessary to make assumptions about the real-life problems in order to
create a model.xy
O(x1, y1)(x2, y2)
x2 – x 1y2 – y 1
x1 cmy
Oy = mx + cSummary of key points
|
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] |
113
Circles
After completing this unit you should be able to:
● Find the mid point of a line segment → pages 114 – 115
● Find the equation of the perpendicular bisector t o a line segment
→ pages 116 – 117
● Know how to find the equation of a circle → pages 117 – 120
● Solve geometric problems involving straight lines and circles
→ pages 121 – 122
● Use circle properties to solve problems on coordinate grids
→ pages 123 – 128
● Find the angle in a semicircl e and solve other problems involving
circles and triangles → pages 128 – 132Objectives
1 Write each of the following in the form
(x + p)2 + q:
a x2 + 10 x + 28 b x2 − 6x + 1
c x2 − 12 x d x2 + 7x ← Section 2.2
2 Find the equation of the line pa ssing
through each of the following pairs of points:
a
A(0, −6) and B(4, 3)
b P(7, −5) and Q(−9, 3)
c R(−4, −2) and T(5, 10) ← Section 5.2
3 Use the discriminant to determine
whether the f
ollowing have two real
solutions, one real solution or no real solutions.
a
x2 − 7x + 14 = 0
b x2 + 11 x + 8 = 0
c 4x2 + 12 x + 9 = 0 ← Section 2.5
4 Find the equation of the line that pa sses
through the point (3, − 4) and is perpendicular
to the line with equation 6x – 5y – 1 = 0
← Section 5.3Prior knowledge check
Geostationary orbits are circular orbits
around the Earth. Meteorologists use geostationary satellites to provide information about the Earth’s surface and atmosphere.6
|
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] |
114
Chapter 6
6.1 Midpoints and perpendicular bisectors
You can find the midpoint of a line segment by averaging the
x- and y-coordinates of its endpoints.
■ The midpoint of a line segment with endpoints ( x1, y1) and ( x2, y2)
is ( x 1 + x 2 _______ 2 , y 1 + y 2 _______ 2 ) .
Example 1
Example 2The line segment AB is a diameter of a circle, where A and B are (−3, 8) and (5, 4) respectively.
Find the coordinates of the centre of the circle.
The line segment PQ is a diameter of the circle centre (2, −2). Given that P is (8, −5), find the
coordinates of Q.OB(5, /four.ss01)(1, 6)A(–3, 8)
xy
The centre of the circle is ( −3 + 5 _______ 2 , 8 + 4 ______ 2 )
= ( 2 __ 2 , 12 ___ 2 ) = (1, 6)
(2, –2)
P(8, –5)Q(a, b)
xy
O
Let Q have coordinates ( a, b).
( 8 + a ______ 2 , −5 +
b _______ 2 ) = (2, − 2)
So 8 +
a ______ 2 = 2 −5 +
b _______ 2 = −2
8 + a = 4 −5 +
b = −4
a = −4 b =
1
So, Q is ( −4, 1).Draw a sketch.
Remember the centre of a circle is the midpoint of
a diameter.
(2, −2) is the mid-point of (a , b) and (8, − 5).
Compare the x - and y -coordinates separately.
Rearrange the equations to find a and b .Use ( x1 + x2 _______ 2 , y1 + y2 ______ 2 ) .
Here (x1, y1) = (−3, 8) and (x2, y2) = (5, 4).
Use ( x1 + x2 _______ 2 , y1 + y2 ______ 2 ) .
Here (x1, y1) = (8, − 5) and (x2, y2) = (a, b).y
x(x1, y1)(x2, y2)
,x1 + x2
2()
Oy1 + y2
2
A line s egment is a finite part of
a straight line with two distinct endpoints.Notation
In coordinate geometry problems, it is often
helpful to draw a sketch showing the information given in the question.Problem-solving
|
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] |
115Circles
1 Find the midpoint of the line segment joining each pair of points:
a (4, 2), (6, 8) b (0, 6), (12, 2) c (2, 2), (−4, 6)
d (−
6, 4), (6, −4) e (7, −4), (
−3, 6) f (−
5, −5), (−11, 8)
g (6a, 4
b), (2a, −4b) h (−4
u, 0), (3u, −2v) i (a
+ b, 2a − b), (3a − b, −b)
j (4 √ __
2 , 1) (2 √ __
2 , 7) k ( √ __
2 − √ __
3 , 3 √ __
2 + 4 √ __
3 ), (3 √ __
2 + √ __
3 , − √ __
2 + 2 √ __
3 )
2 The line segment AB has endpoints
A(−2, 5) and B(a, b). The midpoint of AB is M(4, 3).
Find the values of a and b.
3 The line segment PQ is a diameter of
a circle, where P and Q are (−4, 6) and (7, 8) respectively.
Find the coordinates of the centre of the circle.
4 The line segment RS is a diameter of
a circle, where R and S
are ( 4a ___ 5 , − 3b ___ 4 ) and ( 2a ___ 5 , 5b ___ 4 ) respectively. Find the coordinates
of the centre of
the circle.
5 The line segment AB is a diameter of
a circle, where A and B are (− 3, −4) and (6, 10) respectively.
a Find the coordinates of
the centre of the circle.
b Show the centre of
the circle lies on the line y = 2x.
6 The line segment JK is a diameter of
a circle, where J and K are ( 3 _ 4 , 4 _ 3 ) and (− 1 _ 2 , 2) respectively.
The centre of the circle lies on the line segment with equation y = 8x + b.
Find the value of b.
7 The line segment AB is a diameter of
a circle, where A and B are (0, −2) and (6, −5) respectively.
Show that the centre of the circle lies on the line x − 2y −10 = 0.
8 The line segment FG is a diameter of
the circle centre (6, 1). Given F is (2, −3), find the
coordinates of G.
9 The line segment CD is a diameter of
the circle centre (−2a, 5a). Given D has coordinates
(3a, −7a), find the coordinates of C.
10 The points M(3,
p) and N(q, 4) lie on the circle centre
(5, 6). The line segment MN is a diameter of the circle. Find the values of p and q.
11
The points V(
−4, 2a) and W(3b, −4) lie on the circle centre (b, 2a). The line segment VW is a
diameter of the circle. Find the values of a and b.P
P
P
P
P
P
P
P
A triangle has vertices at A (3, 5), B (7, 11) and C (p, q). The midpoint of side BC is
M(8, 5).
a Fin
d the values of p and q.
b Fin
d the equation of the straight line joining the midpoint of AB to the
point M .
c Sho
w that the line in part b is parallel to the line AC .ChallengeExercise 6A
Your answer will be in terms of
a and b .Problem-solving
Use the formula for finding the midpoint:
( 3 + q _____ 2 , p +
4 _____ 2 ) = (5, 6)Problem-solving
You can also
pr
ove results like this
using vectors.
→ Sec tion 11.5Links
|
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] |
116
Chapter 6
■ The perpendicular bisector of a line segment AB is the straight line
that is perpendicular to AB and passes through the midpoint of AB.lB
Amidpoint
If the gradient of AB is m then the gradient
of its perpendicular bisector, l , will be − 1 __ m
OB(5, 2)A(–1, /four.ss01)
xy
l
C
The centre of the circle is ( −1 + 5 _______ 2 , 4 + 2 ______ 2 )
= (2,
3)
The gradient of the line segment AB
is 2 − 4 ________ 5 − (−1) = − 1 __ 3
Grad
ient of l = 3.
The equation of l is
y − 3 = 3( x − 2)
y − 3 = 3 x − 6
So y =
3x − 3Example 3
The line segment AB is a diameter of the circle centre C , where A and B are (−1, 4) and (5, 2)
respectively. The line l passes through C and is perpendicular to AB. Find the equation of l.
1 Find the perpendicular bisector of the line segment joining each pair of points:
a A(−
5, 8) and B(7, 2) b C(−
4, 7) and D(2, 25) c E(3, −
3) and F(13, −7)
d P(−
4, 7) and Q(−4, −1) e S(4, 11) and T
(−5, −1) f X(13, 11) and Y
(5, 11)
2 The line FG is a diameter of
the circle centre C, where F and G are (−2, 5) and (2, 9) respectively.
The line l passes through C and is perpendicular to FG. Find the equation of l. (7 marks)
3 The line JK is a diameter of
the circle centre P, where J and K are (0, −3) and (4, −5) respectively.
The line l passes through P and is perpendicular to JK. Find the equation of l. Write your
answer in the form ax + by + c = 0, where a, b and c are integers.
4 Points A
, B, C and D have coordinates A(−4, −9), B(6, −3), C(11, 5) and D(−1, 9).
a Find the equation of the perpendicular bisector of
line segment AB.
b Find the equation of the perpendicular bisector of
line segment CD.
c Find the coordinates of
the point of intersection of the two perpendicular bisectors.E/P
PExercise 6BDraw a sketch.
l is the perpendicular bisector of AB.
Remember the product of the gradients of two
perpendicular lines is = −1, so − 1 __ 3 × 3 = −1.
The perpendicular line l passes through the point
(2, 3) and has gradient 3, so use y − y1 = m (x − x1)
with m = 3 and (x1, y1) = (2, 3).
Rearrange the equation into the form y = mx + c.Use ( x1 + x2 _______ 2 , y1 + y2 ______ 2 ) .
Here (x1, y1) = (−1, 4) and (x2, y2) = (5, 2).
Use m = y2 − y1 ______ x2 − x1 . Here (x1, y1) = (−1, 4) and
(x2, y2) = (5, 2).
|
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] |
117Circles
6.2 Equation of a circle
A circle is the set of points that are equidistant from a fixed point. You can use Pythagoras’ theorem
to derive equations of circles on a coordinate grid.
For any point (x, y) on the circumference of a circle, you can use Pythagoras’ theorem to show
the relationship between x, y and the radius r.
■ The equation of a circl
e with centre (0, 0) and radius r is x2 + y2 = r2.
When a circle has a centre (a, b) and radius r, you can use the following
general form of the equation of a circle.
■ The equation of the circl
e with centre ( a, b) and radius r is
(x − a)2 + ( y − b)2 = r2.y
x Or
(a, b) This point of intersection
is c
alled the circumcentre of the
triangle. → Sec tion 6.5Links5 Point X
has coordinates (7, −2) and point Y has coordinates
(4, q). The perpendicular bisector of XY has equation
y = 4x + b. Find the value of q and the value of b.It is often easier to find unknown
values in the order they are given in the question. Find q first, then find b .Problem-solvingP
Triangle PQR has vertices at P (6, 9), Q (3, − 3) and R (−9, 3).
a Fin
d the perpendicular bisectors of each side of the triangle.
b Sho
w that all three perpendicular bisectors meet at a single
point, and find the coordinates of that point.Challenge
This circle is a translation of the circle
x2 + y2 = r2 by the vector ( a b ) . ← Section 4.5Links
Example 4
Write down the equation of the circle with centre (5, 7) and radius 4.
(x, y)
y – 7
x – 5(5, 7)/four.ss01y
x O
(x − 5)2 + (y − 7)2 = 42
(x − 5)2 + (y − 7)2 = 16 Simplify by calculating 42 = 16.Substitute a = 5, b = 7 and r = 4 into the equation. Explore the general form of the
eq
uation of a circle using GeoGebra.Online
|
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] |
118
Chapter 6
Example 6
The line segment AB is a diameter of a circle, where A and B are (4, 7) and (−8, 3) respectively.
Find the equation of the circle.
Length of AB = √ _____________________ (4 − ( −8))2 + (7 − 3)2
= √ ________ 122 + 42
= √ ____ 160
= √ ___ 16 × √ ___ 10
= 4 √ ___ 10
So th
e radius is 2 √ ___ 10 .
The c
entre is ( 4 + (−8) ________ 2 , 7 + 3 ______ 2 ) = (−2, 5).
The equation of the circle is
(x + 2)2 + (y − 5)2 = (2 √ ___ 10 )2
Or (x + 2)2 + (y − 5)2 = 40.Use d = √ ____________________ (x2 − x1)2 + (y2 − y1)2
Here
(x1, y1) = (−8, 3) and ( x2, y2) = (4, 7)Example 5
A circle has equation (x − 3)2 + (y + 4)2 = 20.
a Write down the centr
e and radius of the circle.
b Show that the cir
cle passes through (5, −8).
a Centre (3, − 4) , radius √ ___ 20 = 2 √ __
5
b (x
− 3)2 + (y + 4)2 = 20
Substitute (5, − 8)
(5 − 3)2 + (−8 + 4)2 = 22 + (−4)2
= 4 + 16
= 20
So the circle passes through the point
(5, −8).Substitute x = 5 and y = − 8 into the equation of
the circle.
(5, − 8) satisfies the equation of the circle.r2 = 20 so r = √ ___ 20
Remember the centre of a circle is at the
midpoint of a diameter. Use ( x1 + x2 _______ 2 , y1 + y2 ______ 2 ) .You need to work out the steps of this problem
yourself: •
Fin
d the radius of the circle by finding the
length of the diameter and dividing by 2.
• Fin
d the centre of the circle by finding the
midpoint of AB .
• Write down the equation of the circle.Problem-solving
You can multiply out the brackets in the equation of a circle to find it in an alternate form:
(x − a)2 + (y − b)2 = r 2
x2 − 2ax + a2 + y2 − 2by + b2 = r 2
x2 + y2 − 2ax − 2by + b2 + a2 − r 2 = 0
■ The equation of a circl e can be given in the form:
x2 + y2 + 2fx + 2gy + c = 0
■ This circle has centre (−f, −g) and radius √ _________ f 2 + g2 − c Compare the constant terms with the
equation given in the key point:b
2 + a2 − r 2 = c so r = √ _________ f 2 + g 2 − c
|
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] |
119Circles
Example 7
Find the centre and the radius of the circle with the equation x2 + y2 − 14x + 16y − 12 = 0.
Rearrange into the form ( x − a)2 + (y − b)2 = r2.
x2 + y2 − 14 x + 16 y − 12 = 0
x2 − 14 x + y2 + 16 y − 12 = 0 (1)
Co
mpleting the square for x terms and y terms.
x2 − 14 x = ( x − 7)2 − 49
y2 + 16 y = ( y + 8)2 − 64
Substituting back into (1)
(x − 7)2 − 49 + ( y + 8)2 − 64 − 12 = 0
(x − 7)2 + (y + 8)2 = 125
(x − 7) 2 + (y + 8) 2 = ( √ ____ 125 ) 2
√ ____ 125 = √ ___ 25 × √ __
5 = 5 √ __
5
The e
quation of the circle is
(x − 7) 2 + (y + 8) 2 = (5 √ __
5 ) 2
The c
ircle has centre (7, − 8) and
radius = 5 √ __
5 .Move the number terms to the right-hand side of
the equation.
Write the equation in the form (x − a )
2 + (y − b )2 = r2.
Simplify √ ____ 125 .
You could also compare the original equation with:
x2 + y2 + 2fx + 2gy + c = 0
f = − 7, g = 8 and c = − 12 so the circle has centre
(7, − 8) and radius √ ______________ (−7) 2 + 8 2 − (−12) = 5 √ __
5 .Group the x terms and y terms together.
1 Write down the equation of each circle:
a Centre (3, 2), radius 4 b Centre (−4, 5), radius 6 c Centre (5, −6), radius 2 √ __
3
d Centre (2a
, 7a), radius 5a e Centre (−
2 √ __
2 , −3 √ __
2 ), radius 1
2 Write down the coor
dinates of the centre and the radius of each circle:
a (x
+ 5)2 + (y − 4)2 = 92 b (x − 7)2 + (y − 1)2 = 16 c (x + 4)2 + y2 = 25
d (x
+ 4a)2 + (y + a)2 = 144a2 e (x − 3 √ __
5 )2 + (y + √ __
5 )2 = 27
3 In each case, show tha
t the circle passes through the given point:
a (x
− 2)2 + (y − 5)2 = 13, point (4, 8) b (x + 7)2 + (y − 2)2 = 65, point (0, −2)
c x2 + y2 = 252, point (7, −24) d (x − 2a)2 + (y + 5a)2 = 20a2, point (6a, −3a)
e (x
− 3 √ __
5 )2 + (y − √ __
5 )2 = (2 √ ___ 10 )2 point, ( √ __
5 , − √ __
5 )
4 The point (4, −2) lies on the cir
cle centre (8, 1).
Find the equation of the circle. First find the radius of the circle. Hint PExercise 6C You need to complete the square fo r the
terms in x and for the terms in y . ← Section 2.2LinksIf you need to find the centre and radius of a circle with an equation given in expanded form it is
usually safest to complete the square for the x and y terms.
|
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] |
120
Chapter 6
5 The line PQ is the diameter of
the circle, where P and Q are (5, 6) and (−2, 2) respectively.
Find the equation of the circle. (5 marks)
6 The point (1, −3) lies on the cir
cle (x − 3)2 + (y + 4)2 = r2. Find the value of r. (3 marks)
7 The points P(2, 2),
Q(2 + √ __
3 , 5) and R(2 − √ __
3 , 5) lie on the circle (x − 2)2 + ( y − 4)2 = r2.
a Find the value of
r. (2 marks)
b Show that
△PQR is equilateral. (3 marks)
8 a Show that
x2 + y2 − 4x − 11 = 0 can be written in the form (x − a)2 + y2 = r2,
where a and r are numbers to be found.
(2 marks)
b Hence write down the centre and r
adius of the circle with
equation x2 + y2 − 4x − 11 = 0 (2 marks)
9 a Show that
x2 + y2 − 10x + 4y − 20 = 0 can be written in the form (x − a)2 + (y − b)2 = r2,
where a, b and r are numbers to be found. (2 marks)
b Hence write down the centre and r
adius of the circle with equation
x2 + y2 − 10x + 4y − 20 = 0. (2 marks)
10 Find the centre and radius of
the circle with each of the following equations.
a x2 + y2 − 2x + 8y − 8 = 0
b x2 + y2 + 12x − 4y = 9
c x2 + y2 − 6y = 22x − 40
d x2 + y2 + 5x − y + 4 = 2y + 8
e 2x2 + 2y2 − 6x + 5y = 2x − 3y − 3
11 A circle C
has equation x2 + y2 + 12x + 2y = k, where k is a constant.
a Find the coordinates of
the centre of C. (2 marks)
b State the range of
possible values of k. (2 marks)
12 The point P(7,
−14) lies on the circle with equation x2 + y2 + 6x − 14y = 17.
The point Q a
lso lies on the circle such that PQ is a diameter.
Find the coordinates of point Q. (4 marks)
13 The circle with equation (
x − k)2 + y2 = 41 passes through the point (3, 4).
Find the two possible values of k. (5 marks)E/P
E/P
E/P
E/P
Start by writing ( x2 – 4x)
in the form ( x – a )2 – b.Problem-solving
E/P
Start by writing the equation
in o
ne of the following forms:
(x − a)2 + (y − b)2 = r2
x2 + y2 + 2fx + 2gy + c = 0Hint
E/P
A circle must have a
positive radius.Problem-solving
E/P
E/P
1 A circle with equation ( x − k)2 + (y − 2)2 = 50 passes through the point (4, − 5).
Find the possible values of k and the equation of each circle.
2 By co
mpleting the square for x and y , show that the equation x2 + y2 + 2fx + 2gy + c = 0
describes a circle with centre ( −f, −g) and radius √ _________ f 2 + g 2 − c .Challenge
|
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] |
121Circles
Solve the equations simultaneously, so substitute
y = x + 5 into the equation of the circle.
← Sec tion 3.26.3 Intersections of straight lines and circles
You can use algebra to find the coordinates of intersection
of a straight line and a circle.
■ A straight line can intersect a cir
cle once, by just
touching the circle, or twice.
Not all straight lines will intersect a given circle.
two points of
intersectionone point of
intersectionno points of
intersectiony
O x
Example 8
Find the coordinates of the points where the line y = x + 5 meets the circle x
2 + (y − 2)2 = 29.
x2 + (y − 2)2 = 29
x2 + (x + 5 − 2)2 = 29
x2 + (x + 3)2 = 29
x2 + x2 + 6 x + 9 = 29
2x2 + 6 x − 20 = 0
x2 + 3 x − 10 = 0
(x + 5)( x − 2) = 0
So x = − 5 and x = 2.
x =
−5: y = − 5 + 5 = 0
x =
2: y = 2 + 5 = 7
The line meets the circle at ( −5, 0) and (2, 7).Simplify the equation to form a quadratic equation.
The resulting quadratic equation has two distinct
solutions, so the line intersects the circle at two distinct points.
Remember to write the answers as coordinates.Now find the y -coordinates, so substitute the
values of x into the equation of the line.
Example 9
Show that the line y = x − 7 does not meet the circle (x + 2)2 + y2 = 33.
(x + 2)2 + y2 = 33
(x + 2)2 + (x − 7)2 = 33
x2 + 4 x + 4 + x2 − 14 x + 49 = 33
2x2 − 10x + 20 = 0
x2 − 5 x + 10 = 0
Now b2 – 4ac = (−5)2 − 4 × 1 × 10
= 25 − 40
= −15
b2 − 4 ac < 0, so the line does not meet the
circle.Try to solve the equations simultaneously, so
substitute y = x − 7 into the equation of the circle.
Use the discriminant b2 − 4ac to test for roots of
the quadratic equation. ← Sec tion 2.5
If b2 − 4ac > 0 there are two distinct roots.
If b2 − 4ac = 0 there is a repeated root.
If b2 − 4ac < 0 there are no real roots.Problem-solving Explore intersections of straight
lin
es and circles using GeoGebra.Online
|
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] |
122
Chapter 6
1 Find the coordinates of
the points where the circle
(x − 1)2 + (y − 3)2 = 45 meets the x-axis. Substitute y = 0 in to the equation. Hint
2 Find the coordinates of
the points where the circle (x − 2)2 + (y + 3)2 = 29 meets the y-axis.
3 The line y =
x + 4 meets the circle (x − 3)2 + (y − 5)2 = 34 at A and B.
Find the coordinates of A and B.
4 Find the coordinates of
the points where the line x + y + 5 = 0 meets the circle
x2 + 6x + y2 + 10y − 31 = 0.
5 Show that the line
x − y − 10 = 0 does not
meet the circle x2 − 4x + y2 = 21.
6 a Show that the line
x + y = 11 meets the circle with equation x2 + (y − 3)2 = 32 at only one
point. (4 marks)
b Find the coordinates of
the point of intersection. (1 mark)
7 The line y =
2x − 2 meets the circle (x − 2)2 + (y − 2)2 = 20 at A and B.
a Find the coordinates of
A and B. (5 marks)
b Show that
AB is a diameter of the circle. (2 marks)
8 The line x + y = a meets the circle (
x − p)2 + (y − 6)2 = 20 at (3, 10), where a and p are constants.
a Work out the v
alue of a. (1 mark)
b Work out the tw
o possible values of p. (5 marks)
9 The circle with equation (
x − 4)2 + (y + 7)2 = 50 meets the straight line with equation
x − y − 5 = 0 at points A and B.
a Find the coordinates of
the points A and B. (5 marks)
b Find the equation of the perpendicular bisector of
line segment AB. (3 marks)
c Show that the perpendicular bisector of
AB passes through the centre of the circle. (1 mark)
d Find the area of triangle
OAB . (2 marks)
10 The line with equation y
= kx intersects the circle with equation x2 − 10x + y2 − 12y + 57 = 0 at
two distinct points. Find a range of possible values of k . Round your answer to 2 decimal places.
a Show that 21
k2 − 60k + 32 < 0. (5 marks)
b Hence determine the range of
possible values for k. (3 marks)
11 The line with equation y
= 4x − 1 does not intersect
the circle with equation x2 + 2x + y2 = k. Find the
range of possible values of k. (8 marks)
12 The line with equation y
= 2x + 5 meets the circle
with equation x2 + kx + y2 = 4 at exactly one point.
Find two possible values of k. (7 marks)PAttempt to solve the equations simultaneously.
Use the discriminant to show that the resulting quadratic equation has no solutions.Problem-solving
E/P
E/P
E/P
E/P
E/P
E/P
E/PExercise 6D
If you are solving a problem where there are 0, 1 or 2 solutions (or points of intersection), you might be able to use the discriminant.Problem-solving
|
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] |
123Circles
6.4 Use tangent and chord pr operties
You can use the properties of tangents and chords within circles to solve
geometric problems. A tangent to a circle is a straight line that intersects the circle at only one point.
■
A tangent to a cir
cle is perpendicular to the radius of the circle at
the point of intersection.
A chord is a line segment that joins two points on the
circumference of a circle.
■ The perpendicular bisector of a chor
d will go through the
centre of a circle.tangentradius
chordcent re
of circl eperpendicular
bisector
Example 10
The circle C has equation (x − 2)2 + (y − 6)2 = 100.
a Verify that the point
P(10, 0) lies on C.
b Find an equation of the tangent to
C at the point (10, 0), giving your answer in the form
ax + by + c = 0.
a (x − 2)2 + (y − 6)2 = (10 − 2)2 + (0 − 6)2
= 82 + (−6)2
= 6
4 + 36
= 100 ✓
b The c
entre of circle C is (2, 6). Find the
gradient of the line between (2, 6) and P .
m = y 2 − y 1 _______ x 2 − x 1 = 6 − 0 _______ 2 − 10 = −6 ___ 8 = − 3 __ 4
The g
radient of the tangent is 4 __ 3
y −
y1 = m(x − x1)
y − 0 = 4 __ 3 (x − 10)
3y =
4x − 40
4x − 3 y − 40 = 0Substitute ( x, y) = (10, 0) into the equation for the
circle.
The point P (10, 0) satisfies the equation, so P lies
on C .
A circle with equation ( x − a )2 + (y − b )2 = r2 has
centre ( a, b).
Use the gradient formula with ( x1, y1) = (10, 0)
and ( x2, y2) = (2, 6)
Substitute ( x1, y1) = (10, 0) and m = 4 _ 3 into the
eq
uation for a straight line.
Simplify.The tangent is perpendicular to the radius at that
point. If the gradient of the radius is m then the
gradient of the tangent will be − 1 __ m
Leave the answer in the correct form. Explore the circle theorems
us
ing GeoGebra.Online
|
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] |
124
Chapter 6
Draw a sketch showing the circle and the two
possible tangents with gradient −3. If you are solving a problem involving tangents and circles there is a good chance you will need to use the radius at the point of intersection, so draw this on your sketch.Problem-solvingExample 11
A circle C has equation (x − 5)2 + (y + 3)2 = 10.
The line l is a tangent to the circle and has gradient −3.
Find two possible equations for l, giving your answers in the form y = mx + c.
(5, –3)x Oy
Find a line that passes through the centre of
the circle that is perpendicular to the tangents.
The gradient of this line is 1 __ 3
The c
oordinates of the centre of circle are
(5, −3)
y − y1 = m(x − x1)
y + 3 =
1 __ 3 (x − 5)
y + 3 =
1 __ 3 x − 5 __ 3
y =
1 __ 3 x − 14 ___ 3
(x −
5)2 + (y + 3)2 = 10
(x − 5)2 + ( 1 __ 3 x − 14 ___ 3 + 3) 2
= 10
(x − 5)2 + ( 1 __ 3 x − 5 __ 3 ) 2
= 10
x 2 − 10x + 25 + 1 __ 9 x 2 − 10 ___ 9 x + 25 ___ 9 = 10
10 ___ 9 x 2 − 100 _____ 9 x + 250 _____ 9 = 10
10x2 − 100x + 250 = 90
10x2 − 100x + 160 = 0
x2 − 10x + 16 = 0
(x − 8)( x − 2) = 0
x = 8 or x = 2
y = − 6 __ 3 = −2 or y = −4The gradient of the tangents is − 3, so the
gradient of a perpendicular line will be −1 ___ −3 = 1 __ 3 This line will intersect the circle at the same
points where the tangent intersects the circle.
A circle with equation ( x − a )2 + (y − b )2 = r2 has
centre ( a, b).
Substitute ( x1, y1) = (5, − 3) and m = 1 _ 3 into the
eq
uation for a straight line.
This is the equation of the line passing through
the circle.
Substitute y = 1 __ 3 x − 14 ___ 3 into the equation for a
circle to find the points of intersection.
Simplify the expression.
Factorise to find the values of x .
Substitute x = 8 and x = 2 into y = 1 __ 3 x − 14 ___ 3 .
|
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] |
125Circles
This is the other possible equation for the tangent. So the tangents will intersect the circle at
(8, −2) and (2, − 4)
y − y1 = m(x − x1)
y + 2 = − 3(x − 8)
y = −3
x + 22
y −
y1 = m(x − x1)
y + 4 = − 3(x − 2)
y = −3
x + 2 This is one possible equation for the tangent.
Substitute ( x 1 , y 1 ) = (2, −4) and m = − 3 in to the
equation for a straight line.Substitute ( x1, y1) = (8, − 2) and m = − 3 into the
equation for a straight line.
Example 12
The points P and Q lie on a circle with centre C, as shown in the diagram.
The point P has coordinates (−7, −1) and the point Q has coordinates (3, −5).M is the midpoint of the line segment PQ.The line l passes through the points M and C.
a
Find an equation for
l.
Given that the y-coordinate of C is −8,b
show that the
x-coordinate of C is −4
c find an equation of the cir
cle.y
x
Q(3, –5)O P(–7, –1)
lCM
a The midpoint M of line segment PQ is:
( x 1 + x 2 ______ 2 , y 1 + y 2 ______ 2 ) = ( −7 + 3 ______ 2 , −1 + (−5) ________ 2 )
= (−2, − 3)
Grad
ient of PQ = y 2 − y 1 ______ x 2 − x 1 = −5 − (−1) ________ 3 − (−7)
= −4 ___ 10 = − 2 __ 5
The g
radient of a line perpendicular to
PQ is 5 __ 2
y −
y1 = m(x − x1)
y +
3 = 5 __ 2 (x + 2)
y +
3 = 5 __ 2 x + 5
y = 5 __ 2 x + 2
b y =
5 __ 2 x + 2
−8 =
5 __ 2 x + 2
5 __ 2 x = −10
x =
−4Use the midpoint formula with ( x1, y1) = (−7, −1)
and ( x1, y1) = (3, − 5).
Use the gradient formula with ( x1, y1) = (−7, −1)
and ( x1, y1) = (3, − 5).
Substitute ( x1, y1) = (−2, −3) and m = 5 _ 2 into the
eq
uation of a straight line.
Simplify and leave in the form y = mx + c.If a gradient is given as a fraction, you can find
the perpendicular gradient quickly by turning the fraction upside down and changing the sign.Problem-solving
The perpendicular bisector of any chord passes through the centre of the circle. Substitute y = − 8
into the equation of the straight line to find the corresponding x -coordinate.
Solve the equation to find x .
|
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] |
126
Chapter 6
c The centre of the circle is ( −4, −8).
To find the radius of the circle:
CQ =
√ ___________________ ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2
= √ _______________________ (3 − (− 4)) 2 + (−5 − (−8)) 2
= √ _______ 49 + 9 = √ ___ 58
So th
e circle has a radius of √ ___ 58 .
The e
quation of the circle is:
(x − a)2 + (y − b)2 = r2
(x + 4)2 + (y + 8)2 = 58The radius is the length of the line segment CP
or CQ .
Substitute ( x1, y1) = (−4, −8) and ( x2, y2) = (3, −5).
Substitute (a, b) = (−4, −8) and r = √ ___ 58 into the
equation of a cir
cle.
Exercise 6E
1 The line x + 3y − 11 = 0 touches the circle (x + 1)2 + (y + 6)2 = r2 at (2, 3).
a Find the radius of the cir
cle.
b Show that the r
adius at (2, 3) is perpendicular to the line.
2 The point P(1,
−2) lies on the circle centre (4, 6).
a Find the equation of the cir
cle.
b Find the equation of the tangent to the cir
cle at P.
3 The points A and
B with coordinates (−1, −9) and (7, −5) lie on the circle C with equation
(x − 1)2 + (y + 3)2 = 40.
a Find the equation of the perpendicular bisector of
the line segment AB.
b Show that the perpendicular of
bisector AB passes through the centre of the circle C.
4 The points P and
Q with coordinates (3, 1) and (5, −3) lie on the circle C with equation
x2 − 4x + y2 + 4y = 2.
a Find the equation of the perpendicular bisector of
the line segment PQ.
b Show that the perpendicular bisector of
PQ passes through the centre of the circle C.
5 The circle C
has equation x2 + 18x + y2 − 2y + 29 = 0.
a Verify the point P
(−7, −6) lies on C. (2 marks)
b Find an equation for the tangent to
C at the
point P, giving your answer in the form
y = mx + b. (4 marks)
c Find the coordinates of
R, the point of intersection of the tangent and the y-axis. (2 marks)
d Find the area of the triangle
APR. (2 marks)P
Ey
xA
O
P(–7, –6)CR
l
|
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] |
127Circles
6 The tangent to the circle (x + 4)2 + (y − 1)2 = 242 at (7, − 10) meets the y -axis at S and the x -axis at T .
a Find the coordinates of
S and T. (5 marks)
b Hence, find the area of
△OST, where O is the origin. (3 marks)
7 The circle C
has equation (x + 5)2 + (y + 3)2 = 80.
The line l is a tangent to the circle and has gradient 2.
Find two possible equations for l giving your answers in
the form y = mx + c. (8 marks)
8 The line with equation 2x
+ y − 5 = 0 is a
tangent to the circle with equation
(x
− 3)2 + (y − p)2 = 5
a Find the two possible v
alues of p. (8 marks)
b Write down the coor
dinates of the centre
of the circle in each case. (2 marks)
9 The circle C
has centre P(11, −5) and passes through the
point Q(5, 3).
a Find an equation for
C. (3 marks)
The line l1 is a tangent to C at the point Q.
b Find an equation for
l1. (4 marks)
The line l2 is parallel to l1 and passes through the midpoint
of PQ. Given that l2 intersects C at A and B
c find the coordinates of
points A and B (4 marks)
d find the length of the line segment AB
, leaving your
answer in its simplest surd form.
(3 marks)
10 The points R and
S lie on a circle with centre C(a, −2),
as shown in the diagram.The point R has coordinates (2, 3) and the point S
has coordinates (10, 1).
M is the midpoint of the line segment RS. The line l passes through M and C. a
Find an equation for
l.
(4 marks)
b Find the value of
a.
(2 marks)
c Find the equation of the cir
cle. (3 marks)
d Find the points of intersection, A
and B, of the line l and the circle. (5 marks)E/P
l1
l2y
x OE/P
E/P
The line is a tangent to the circle so it must
intersect at exactly one point. You can use the discriminant to determine the values of p for which this occurs.Problem-solving
E/Pl1
l2BQ(5, 3)
P(11, –5)
CAy
O x
E/P
R(2, 3)
S(10, 1)A
C(a, –2)
By
O xM
|
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] |
128
Chapter 6
11 The circle C
has equation x2 − 4x + y2 − 6y = 7.
The line l with equation x − 3y + 17 = 0 intersects the circle
at the points P and Q.
a Find the coordinates of
the point P and the
point Q. (4 marks)
b Find the equation of the tangent a
t the point P
and the point Q. (4 marks)
c Find the equation of the perpendicular bisector
of
the chord PQ. (3 marks)
d Show that the tw
o tangents and the perpendicular bisector intersect at a single point
and find the coordinates of the point of intersection. (2 marks)E/P
P
OQ
Cly
x
Use the point (0, −2) to write
an equation for the tangent in terms of m. Substitute this equation into the equation for the circle.Problem-solving
1 The circle C has equation ( x − 7)2 + (y + 1)2 = 5.
The line l with positive gradient passes through (0, − 2) and is a
tangent to the circle.
Find an equation of l , giving your answer in the form y = mx + c.
l
C –2y
O x
2 The circle with centre C has equation ( x − 2)2 + (y − 1)2 = 10.
The tangents to the circle at points P and Q meet at the point R with
coordinates (6, − 1).
a Sho
w that CPRQ is a square.
b Hen
ce find the equations of both tangents.Challenge
6.5 Circles and triangles
A triangle consists of three points, called vertices.
It is always possible to draw a unique circle through the three vertices of any triangle. This circle is called the circumcircle of
the triangle. The centre of the circle is called the circumcentre of the triangle and is the point where the perpendicular bisectors of each side intersect.y
xCircumcircle
|
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] |
129Circles
Example 13
The points A(−8, 1), B(4, 5) and C(−4, 9) lie on the
circle, as shown in the diagram.
a Show that
AB is a diameter of the circle.
b Find an equation of the cir
cle.y
xC(–4, 9)
OA(–8, 1)B(4, 5)
a Test triangle ABC to s ee if it is a right-
angled triangle.
AB2 = (4−(−8))2 + (5 − 1)2
= 122 + 42 = 160
AC2 = (−4−(−8))2 + (9 − 1)2
= 42 + 82 = 80
BC2 = (−4 − 4)2 + (9 − 5)2
= (−8)2 + 42 = 80
Now, 80 + 80 = 160 so AC2 + BC2 = AB2
So triangle ABC is a right-angled triangle
and AB is the diameter of the circle.
b Fin
d the midpoint M of AB .
( x 1 + x 2 ______ 2 , y 1 + y 2 ______ 2 ) = ( −8 + 4 ______ 2 , 1 + 5 _____ 2 )
= (−2, 3)
The d
iameter is √ ____ 160 = 4 √ ___ 10
The r
adius is 2 √ ___ 10
(x −
a)2 + (y − b)2 = r2
(x + 2)2 + (y − 3)2 = (2 √ ___ 10 ) 2
(x + 2)2 + (y − 3)2 = 40Use d2 = (x2 − x1)2 + (y2 − y1)2 to determine the
length of each side of the triangle ABC .
Use Pythagoras’ theorem to test if triangle ABC
is a right-angled triangle.
If ABC is a right-angled triangle, its longest
side must be a diameter of the circle that passes
through all three points.For a right-angled triangle, the hypotenuse of the triangle is a
diameter of the circumcircle.
You can state this result in two other ways:
■ If ∠ PRQ =
90° then R lies on the circle with diameter PQ.
■ The angle in a semicircl
e is always a right angle.
To find the centre of a circle given any three points on the circumference:■
Find the equations of the perpendicular bisectors
o
f two different chords.
■ Find the coordinat
es of the point of intersection of
the perpendicular bisectors.PQR
Perpendicular
bisectors intersectat the cent re of
the circle.
The centre of the circle is the midpoint of AB .
Substitute ( x1, y1) = (−8, 1) and ( x2, y2) = (4, 5).
From part a , A B 2 = 160 .
The radius is half the diameter.
Substitute ( a, b) = (−2, 3) and r = 2 √ ___ 10 into the
eq
uation for a circle.
|
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] |
130
Chapter 6
Example 14
The points P(3, 16), Q(11, 12) and R(−7, 6) lie on the circumference of a circle. The equation of the
perpendicular bisector of PQ is y = 2x.
a Find the equation of the perpendicular bisector of
PR.
b Find the centre of the cir
cle.
c Work out the equa
tion of the circle.
a The midpoint of PR is ( x 1 + x 2 ______ 2 , y 1 + y 2 ______ 2 )
= ( 3 + (−7) _______ 2 , 16 + 6 ______ 2 ) = (−2, 11)
The g
radient of PR is
y2 − y 1 _______ x2 − x 1 = 6 − 16 ______ −7 − 3
= −10 _____ −10 = 1
The g
radient of a line perpendicular to
PR is −1.
y − y1 = m(x − x1)
y − 11 = − 1(x − (−2))
y − 11 = − x − 2
y = − x + 9
b Equ
ation of perpendicular bisector to
PQ: y = 2x
Equation of perpendicular bisector to
PR: y = − x + 9
2x = − x + 9
3x = 9
x = 3y = 2xy = 2(3) = 6
The centre of the circle is at (3, 6).
c
Fin
d the distance between (3, 6) and
Q(11, 12).
d =
√ ____________________ ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2
d =
√ ___________________ (11 − 3) 2 + (12 − 6) 2
d =
√ _________ 64 + 36
d =
√ ____ 100 = 10
The c
ircle through the points P , Q and R
has a radius of 10.
The centre of the circle is (3, 6).The equation for the circle is
(x − 3)
2 + (y − 6)2 = 100The perpendicular bisector of PR passes through
the midpoint of PR .
Substitute ( x1, y1) = (3, 16) and ( x2, y2) = (−7, 6)
into the midpoint formula.
Substitute ( x1, y1) = (3, 16) and ( x2, y2) = (−7, 6)
into the gradient formula.
Simplify and leave in the form y = mx + c.Substitute m = − 1 and ( x1, y1) = (−2, 11) into the
equation for a straight line.
Solve these two equations simultaneously to find
the point of intersection. The two perpendicular bisectors intersect at the centre of the circle.
This is the x -coordinate of the centre of the circle.
Substitute x = 3 to find the y -coordinate of the
centre of the circle.
The radius of the circle is the distance from the centre to a point on the circumference of the circle.
Substitute ( x1, y1) = (3, 6) and ( x2, y2) = (11, 12)
into the distance formula.
Simplify to find the radius of the circle.
Substitute ( a, b) = (3, 6) and r = 10 into
(x − a )2 + (y − b )2 = r2 Explore triangles and
th
eir circumcircles using GeoGebra.Online
|
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] |
131Circles
Exercise 6F
1 The points U( −2, 8), V(7, 7) and W(−3, −1) lie on a circle.
a Show that triangle
UVW has a right angle.
b Find the coordinates of
the centre of the circle.
c Write down an equa
tion for the circle.
2 The points A(2, 6),
B(5, 7) and C(8, −2) lie on a circle.
a Show that
AC is a diameter of the circle.
b Write down an equa
tion for the circle.
c Find the area of the triangle
ABC.
3 The points A(
−3, 19), B(9, 11) and C(−15, 1) lie on the circumference of a circle.
a Find the equation of the perpendicular bisector of
i AB ii AC
b Find the coordinates of
the centre of the circle.
c Write down an equa
tion for the circle.
4 The points P(
−11, 8), Q(−6, −7) and R(4, −7) lie on the circumference of a circle.
a Find the equation of the perpendicular bisector of
i PQ ii QR
b Find an equation for the cir
cle.
5 The points R(
−2, 1), S(4, 3) and T(10, −5) lie on the
circumference of a circle C. Find an equation for the circle.
6 Consider the points A(3, 15),
B(−13, 3), C(−7, −5) and D(8, 0).
a Show that
ABC is a right-angled triangle.
b Find the equation of the cir
cumcircle.
c Hence show that
A, B, C and D all lie on the circumference of this circle.
7 The points A(
−1, 9), B(6, 10), C(7, 3) and D(0, 2) lie on a circle.
a Show that
ABCD is a square.
b Find the area of
ABCD.
c Find the centre of the cir
cle.
8 The points D(
−12, −3), E(−10, b) and F(2, −5) lie
on the circle C as shown in the diagram.
Given that ∠ DEF =
90° and b > 0
a show that
b = 1 (5 marks)
b find an equation for
C. (4 marks)PUse headings in your working
to keep track of what you are working out at each stage.Problem-solving
P
P
E/PE(–10, b)
D(–12, –3)
F(2, –5)y
O x
|
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] |
132
Chapter 6
9 A circle has equation
x2 + 2x + y2 − 24y − 24 = 0
a Find the centre and radius of
the circle. (3 marks)
b The points A(
−13, 17) and B(11, 7) both lie on the circumference of the circle.
Show that AB is a diameter of the circle. (3 marks)
c The point C lies on the nega
tive x-axis and the angle ACB = 90°.
Find the coordinates of C. (3 marks)E/P
1 The line segment QR is a diameter of the circle centre C, where Q and R have coordinates
(11, 12) and (−5, 0) respectively. The point P has coordinates (13, 6).
a Find the coordinates of
C.
b Find the radius of the cir
cle.
c Write down the equa
tion of the circle.
d Show that
P lies on the circle.
2 Show that (0, 0) lies inside the cir
cle (x − 5)2 + (y + 2)2 = 30.
3 The circle C
has equation x2 + 3x + y2 + 6y = 3x − 2y − 7.
a Find the centre and radius of
the circle. (4 marks)
b Find the points of intersection of the cir
cle and the y-axis. (3 marks)
c Show that the cir
cle does not intersect the x-axis. (2 marks)
4 The centres of the cir cles (x − 8)2 + (y − 8)2 = 117 and (x + 1)2 + (y − 3)2 = 106 are P and Q
respectively.
a Show that
P lies on (x + 1)2 + (y − 3)2 = 106.
b Find the length of PQ
.
5 The points A(
−1, 0), B( 1 _ 2 , √ __
3 __ 2 ) and C( 1 _ 2 , − √ __
3 __ 2 ) are the vertices of a triangle.
a Show that the cir
cle x2 + y2 = 1 passes through the vertices of the triangle.
b Show that
△ABC is equilateral.
6 A circle with equation (
x − k)2 + (y − 3k)2 = 13 passes through the point (3, 0).
a Find two possible v
alues of k. (6 marks)
b Given tha
t k > 0, write down the equation of the circle. (1 mark)
7 The line with 3x
− y − 9 = 0 does not intersect the circle with equation x2 + px + y2 + 4y = 20.
Show that 42 − √ ___ 10 < p < 42 + 10 √ ___ 10 . (6 marks)
8 The line y =
2x − 8 meets the coordinate axes at A and B. The line segment AB is a diameter
of the circle. Find the equation of the circle.
9 The circle centre (8, 10) meets the
x-axis at (4, 0) and (a, 0).
a Find the radius of the cir
cle.
b Find the value of
a.P
P
E/P
P
E/P
E/P
P
PMixed exercise 6
|
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] |
133Circles
10 The circle (x − 5)2 + y2 = 36 meets the x-axis at P and Q. Find the coordinates of P and Q.
11 The circle (x
+ 4)2 + (y − 7)2 = 121 meets the y-axis at (0, m) and (0, n).
Find the values of m and n.
12 The circle C
with equation (x + 5)2 + (y + 2)2 = 125 meets the positive coordinate axes at
A(a, 0) and B(0, b).
a Find the values of
a and b. (2 marks)
b Find the equation of the line
AB. (2 marks)
c Find the area of the triangle
OAB , where O is the origin. (2 marks)
13 The circle, centr
e (p, q) radius 25, meets the x-axis at (−7, 0) and (7, 0), where q > 0.
a Find the values of
p and q.
b Find the coordinates of
the points where the circle meets the y-axis.
14 The point A(
−3, −7) lies on the circle centre (5, 1). Find the equation of the tangent to the
circle at A.
15 The line segment AB is a chor
d of a circle centre (2, −1), where A and B are (3, 7) and (−5, 3)
respectively. AC is a diameter of the circle. Find the area of △ABC.
16 The circle C has equation (x − 6)2 + (y − 5)2 = 17.
The lines l1 and l2 are each a tangent to the circle
and intersect at the point (0, 12).
Find the equations of l1 and l2, giving your
answers in the form y = mx + c. (8 marks)
17 The points A and
B lie on a circle with centre C,
as shown in the diagram.The point A has coordinates (3, 7) and the
point B has coordinates (5, 1).
M is the midpoint of the line segment AB.The line l passes through the points M and C.
a
Find an equation for
l. (4 marks)
Giv
en that the x-coordinate of C is −2:
b find an equation of the cir
cle (4 marks)
c find th e area of the triangle ABC. (3 marks)E
P
P
P
E/P
l2
l112y
O x
E/P
lCMA (3, 7)
B (5, 1)
x Oy
|
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] |
134
Chapter 6
18 The circle C
has equation (x − 3)2 + (y + 3)2 = 52.
The baselines l1 and l2 are tangents to the circle and
have gradient 3 _ 2
a Find the points of intersection, P and
Q, of the
tangents and the circle. (6 marks)
b Find the equations of lines
l1 and l2, giving your
answers in the form ax + by + c = 0. (2 marks)
19 The circle C
has equation x2 + 6x + y2 − 2y = 7.
The lines l1 and l2 are tangents to the circle.
They intersect at the point R(0, 6).
a Find the equations of lines
l1 and l2, giving your
answers in the form y = mx + b. (6 marks)
b Find the points of intersection, P
and Q, of the
tangents and the circle. (4 marks)
c Find the area of quadrila
teral APRQ. (2 marks)
20 The circle C
has a centre at (6, 9) and a radius of √ ___ 50 .
The line l1 with equation x + y − 21 = 0 intersects the circle at the points P and Q.
a Find the coordinates of
the point P and the
point Q. (5 marks)
b Find the equations of
l2 and l3, the tangents at the
points P and Q respectively. (4 marks)
c Find the equation of l4, the perpendicular bisector
of the chord PQ. (4 marks)
d Show that the tw
o tangents and the perpendicular
bisector intersect and find the coordinates of R,
the point of intersection. (2 marks)
e Calculate the ar
ea of the kite APRQ. (3 marks)
21 The line y = −
3x + 12 meets the coordinate axes at A and B.
a Find the coordinates of
A and B.
b Find the coordinates of
the midpoint of AB.
c Find the equation of the cir
cle that passes through A, B and O, where O is the origin.
22 The points A(
−3, −2), B(−6, 0) and C(1, q) lie on the circumference of a circle such that
∠ BAC =
90°.
a Find the value of
q. (4 marks)
b Find the equation of the cir
cle. (4 marks)E/P
l1l2y
O x
E/P
P R(0, 6)
CA
OQl2l1y
x
E/P
l3l4l2l1
A
OQPR
xy
P
E/P
|
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] |
135Circles
23 The points R(
−4, 3), S(7, 4) and T(8, −7) lie on the circumference of a circle.
a Show that
RT is the diameter of the circle. (4 marks)
b Find the equation of the cir
cle. (4 marks)
24 The points A(
−4, 0), B(4, 8) and C(6, 0) lie on the circumference of circle C.
Find the equation of the circle.
25 The points A(
−7, 7), B(1, 9), C(3, 1) and D(−7, 1) lie on a circle.
a Find the equation of the perpendicular bisector of:
i AB ii CD
b Find the equation of the cir
cle.
The circle with equation ( x − 5)2 + (y − 3)2 = 20 with centre A
intersects the circle with equation ( x − 10)2 + (y − 8)2 = 10 with
centre B at the points P and Q .
a Fin
d the equation of the line containing the points P and Q
in the form ax + by + c = 0.
b Fin
d the coordinates of the points P and Q .
c Fin
d the area of the kite APBQ .A
OQBP
xy ChallengeE/P
P
P
1 The midpoint of a line segment with endpoints
(x1, y1) and (x2, y2) is ( x 1 + x 2 ______ 2 , y 1 + y 2 ______ 2 ) .
2 The perpendicular bisector of a line segment
AB is the straight line that is perpendicular to
AB and passes through the midpoint of AB.
lB
Amidpointy
x(x1, y1)(x2, y2)
,x1 + x2
2()
Oy1 + y2
2
If the gradient of AB is m
then the gradient of its
perpendicular bisector, l ,
will be − 1 __ m .Summary of key points
|
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] |
136
Chapter 6
3 The equation of a circl e with centre (0, 0) and radius r is x2 + y2 = r2.
4 The equation of the circl
e with centre (a, b) and radius r is (x − a)2 + (y − b)2 = r2.
5 The equation of a circl
e can be given in the form: x2 + y2 + 2fx + 2gy + c = 0
This circle has centre (−f, −g) and radius √ _________ f 2 + g2 − c
6 A straight line can int
ersect a circle once, by just touching
the circle, or twice. Not all straight lines will intersect a
given circle.
7 A tangent to a cir
cle is perpendicular to the radius of the circle at the
point of intersection.
8 The perpendicular bisector of a chor
d will go through the
centre of a circle.
9 • If ∠ PRQ =
90° then R lies on the circle with diameter PQ.
• The angle in a semicircl
e is always a right angle.
10 To find the cent
re of a circle given any three points:
• Find the equations of the perpendicular bisect
ors of two different chords.
• Find the coordinat
es of intersection of the
perpendicular bisectors.two points of
intersectionone point of
intersectionno points of
intersectiony
x
tangentradius
chordcent re
of circl eperpendicular
bisector
PQR
Perpendicular
bisectors intersectat the cent re of
the circle
|
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] |
137
Algebraic methods
After completing this unit you should be able to:
● Cancel fact
ors in algebraic fractions → pages 138–139
● Divide a polynomial by a linear expr ession → pages 139–142
● Use the factor theorem t o factorise a cubic expression
→ pages 143–146
● Construct mathematical proofs using algebra → pages 146–150
● Use proof by exhaustion and disproof by
count
er-example → pages 150–152Objectives
1 Simplify:
a 3x2 × 5x5 b 5x3y2 ______ 15x2y3 ← Section 1.1
2 Factorise:
a x2 − 2x − 24 b 3x2 − 17x + 20
← Sec tion 1.3
3 Use long division to calculate:a
197 041 ÷ 23 b 56 168 ÷ 34
← GC SE Mathematics
4 Find the equations of the lines that pa ss
through these pairs of points:a
(−
1, 4) and (5, −14)
b (2, −6) and (8,
−3) ← GCSE Mathematics
5 Complete the square for the expressions:a
x2 − 2x − 20 b 2x2 + 4x + 15
← Sec tion 2.2Prior knowledge check
Proof is the cornerstone of mathematics.
Mathematicians need to prove theorems (such as Pythagoras’ theorem) before they can use them to solve problems. Pythagoras' theorem can be use to find approximate values for π.
→ Mixed exercise challenge Q17
|
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] |
138
Chapter 7
7.1 Algebraic fractions
You can simplify algebraic fractions using division.
■ When simplifying an algebr
aic fraction, where
possible factorise the numerator and denominator
and then cancel common factors.5x2 – 245
2x2 – 15x + 7=5(x + 7)(x – 7)
(2x – 1)(x – 7)=5(x + 7)
2x – 1Factorise
Cancel common factor
Example 1
Simplify these fractions:
a 7x4 − 2x3 + 6x _____________ x b (x
+ 7)(2x − 1) _____________ (2x − 1) c x2 + 7x + 12 ___________ (x + 3) d x2 + 6x + 5 ___________ x2 + 3x − 10 e 2x2 + 11x + 12 _____________ (x + 3)(x + 4)
Exercise 7Aa 7x4 − 2x3 + 6 x _______________ x
= 7x4 ____ x − 2x3 ____ x + 6x ___ x
= 7x3 − 2x2 + 6
b (x +
7)(2 x − 1) ______________ (2x − 1) = x + 7
c x2 + 7x + 12 _____________ (x + 3) = (x
+ 3)(x + 4) _____________ (x + 3)
= x
+ 4
d x2 + 6 x + 5 _____________ x2 + 3 x − 10 = (x
+ 5)(x + 1) _____________ (x + 5)(x − 2)
= x +
1 ______ x − 2
e 2x2 + 11x + 12 = 2x2 + 3x + 8x + 12
= x(2x + 3) + 4(2x + 3)
= (2x + 3)(x + 4)
So 2x2 + 11x + 12 _______________ (x + 3)(x + 4)
= (2x
+ 3)(x + 4) ______________ (x + 3)(x + 4)
= 2x
+ 3 _______ x + 3 Divide each numerator by x.
Factorise:
x2 + 7x + 12 = (x + 3)(x + 4).Simplify by cancelling the common factor of (2x − 1).
Cancel the common factor of (x + 3).
Cancel the common factor of (x + 5).
Factorise:2x
2 + 11x + 12 = 2x2 + 3x + 8x + 12
1 Simplify these fractions:
a 4x4 + 5x2 − 7x _____________ x b 7x5 − 5x5 + 9x3 + x2 _________________ x c −x4 + 4x2 + 6 ____________ x Factorise: x2 + 6x + 5 = (x + 5)(x + 1) and
x2 + 3x − 10 = (x + 5)(x − 2).
Cancel the common factor of (x + 4).
|
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] |
139
Algebraic methods
d 7x5 − x3 − 4 ___________ x e 8x4 − 4x3 + 6x _____________ 2x f 9x2 − 12x3 − 3x ______________ 3x
g 7x3 − x4 − 2 ___________ 5x h −4x2 + 6x4 − 2x ______________ −2x i −x8 + 9x4 − 4x3 + 6 _________________ −2x
j −9x9 − 6x6 + 4x4 − 2 __________________ −3x
2 Simplify these fractions as far as possible:
a (x
+ 3)(x − 2) ____________ (x − 2) b (x
+ 4)(3x − 1) _____________ (3x − 1) c (x
+ 3)2 _______ (x + 3)
d x2 + 10x + 21 ____________ (x + 3) e x2 + 9x + 20 ___________ (x + 4) f x2 + x − 12 __________ (x − 3)
g x2 + x − 20 ___________ x2 + 2x − 15 h x2 + 3x + 2 __________ x2 + 5x + 4 i x2 + x − 12 ___________ x2 − 9x + 18
j 2x2 + 7x + 6 ____________ (x − 5)(x + 2) k 2x2 + 9x − 18 ____________ (x + 6)(x + 1) l 3x2 − 7x + 2 _____________ (3x − 1)(x + 2)
m 2x2 + 3x + 1 ___________ x2 − x − 2 n x2 + 6x + 8 ___________ 3x2 + 7x + 2 o 2x2 − 5x − 3 ___________ 2x2 − 9x + 9
3 6x3 + 3x2 − 84x ______________ 6x2 − 33x + 42 = ax(x
+ b) ________ x + c , where a, b and c are constants.
Work out the values of a, b and c. (4 marks)E/P
7.2 Dividing polynomials
A polynomial is a finite expression with positive
whole number indices.
■ You can use long division t
o divide a polynomial
by (x ± p), where p is a constant.Polynomials Not polynomials
2x + 4 √ __
x
4xy2 + 3x − 9 6x−2
8 4 __ x
Example 2
Divide x3 + 2x2 − 17x + 6 by (x − 3).
1 x2
x 3 + 2 x 2 − 17 x + 6 ____ x − 3)
x3 − 3 x2
5x2 − 17 xStart by dividing the first term of the polynomial
by x, so that x3 ÷ x = x2.
Next multiply (x − 3) by x2, so that
x2 × (x − 3) = x3 − 3x2.
Now subtract, so that (x3 + 2x2) − (x3 − 3x2) = 5x2.
Finally copy −17x.
|
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] |
140
Chapter 7
2 x2 + 5 x
x 3 + 2 x 2 − 17 x + 6 ____ x − 3)
x3 − 3 x2
5x2 − 17 x
5x2 − 15 x
−2x + 6
3 x2 + 5 x − 2
x 3 + 2 x 2 − 17 x + 6 ____ x − 3)
x3 − 3 x2
5x2 − 17 x
5x2 − 15 x
−2x + 6
−2x + 6
0
So x3 + 2x2 − 17 x + 6 __________________ x − 3 = x2 + 5 x − 2Repeat the method. Divide 5x2 by x, so that
5x2 ÷ x = 5x.
Multiply (x − 3) by 5x, so that
5x × (x − 3) = 5x2 − 15x.
Subtract, so that (5x2 − 17x) − (5x2 − 15x) = −2x.
Copy 6.
Repeat the method. Divide −2x by x, so that −2x ÷ x = −2.
Multiply (x − 3) by −2, so that
−2 × (x − 3) = −2x + 6.
Subtract, so that (−2x + 6) − (−2x + 6) = 0.No numbers left to copy, so you have finished.
This is called the quotient.
(4x4 – 17x2 + 4) ÷ (2x + 1) = 2x3 − x2 − 8x + 4.Example 3
f(x) = 4x4 − 17x2 + 4
Divide f(x) by (2x + 1), giving your answer in the form f(x) = (2x + 1)(ax3 + bx2 + cx + d ).
Find (4 x4 − 17 x2 + 4) ÷ (2 x + 1)
2x3 − x2 − 8 x + 4
4 x 4 + 0 x 3 − 17 x2 + 0 x + 4 ______ 2x + 1 )
4x4 + 2x3
−2x3 − 17 x2
−2x3 − x2
−16x2 + 0 x
−16x2 − 8 x
8x + 4
8x + 4
0
So
4x 4 − 17x2 + 4 = (2x + 1)(2x3 − x2 − 8x + 4). You need to multiply (2x + 1) by 2x3 to get the
4x4 term, so write 2x3 in the answer, and write
2x3(2x + 1) = 4x4 + 2x3 below. Subtract and copy
the next term.
You need to multiply (2x + 1) by –x2 to get the
–2x3 term, so write –x2 in the answer, and write
–x3(2x + 1) = –2x3 – x2 below. Subtract and copy
the next term.
Repeat the method.Use long division. Include the terms 0x3 and 0x
when you write out f(x).
|
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] |
141
Algebraic methods
Example 4
Find the remainder when 2x3 − 5x2 − 16x + 10 is divided by (x − 4).
2x2 + 3 x − 4
2 x 3 − 5 x2 − 16 x + 10 _____ x − 4)
2x3 − 8 x2
3x2 − 16 x
3x2 − 12 x
−4x + 10
−4x + 16
−6
So the remainder is − 6. (x − 4) is not a factor of 2x3 − 5x2 − 16x + 10 as
the remainder ≠ 0.
This means you cannot write the expression in
the form (x − 4)(ax2 + bx + c).
1 Write each polynomial in the form (x ± p)(ax2 + bx + c) by dividing:
a x3 + 6x2 + 8x + 3 by (x + 1) b x3 + 10x2 + 25x + 4 by (x + 4)
c x3 − x2 + x + 14 by (x + 2) d x3 + x2 − 7x − 15 by (x − 3)
e x3 − 8x2 + 13x + 10 by (x − 5) f x3 − 5x2 − 6x − 56 by (x − 7)
2 Write each polynomia
l in the form (x ± p)(ax2 + bx + c) by dividing:
a 6x3 + 27x2 + 14x + 8 by (x + 4) b 4x3 + 9x2 − 3x − 10 by (x + 2)
c 2x3 + 4x2 − 9x − 9 by (x + 3) d 2x3 − 15x2 + 14x + 24 by (x − 6)
e −5x3 − 27x2 + 23x + 30 by (x + 6) f −4x3 + 9x2 − 3x + 2 by (x − 2)
3 Divide:
a x4 + 5x3 + 2x2 − 7x + 2 by (x + 2) b 4x4 + 14x3 + 3x2 − 14x − 15 by (x + 3)
c −3x4 + 9x3 − 10x2 + x + 14 by (x − 2) d −5x5 + 7x4 + 2x3 − 7x2 + 10x − 7 by (x − 1)
4 Divide:a
3x4 + 8x3 − 11x2 + 2x + 8 by (3x + 2) b 4x4 − 3x3 + 11x2 − x − 1 by (4x + 1)
c 4x4 − 6x3 + 10x2 − 11x − 6 by (2x − 3) d 6x5 + 13x4 − 4x3 − 9x2 + 21x + 18 by (2x + 3)
e 6x5 − 8x4 + 11x3 + 9x2 − 25x + 7 by (3x − 1) f 8x5 − 26x4 + 11x3 + 22x2 − 40x + 25 by (2x − 5)
g 25x4 + 75x3 + 6x2 − 28x − 6 by (5x + 3) h 21x5 + 29x4 − 10x3 + 42x − 12 by (7x − 2)
5 Divide:a
x3 + x + 10 by (x + 2) b 2x3 − 17x + 3 by (x + 3)
c −3x3 + 50x − 8 by (x − 4)
6 Divide:a
x3 + x2 − 36 by (x − 3) b 2x3 + 9x2 + 25 by (x + 5)
c −3x3 + 11x2 − 20 by (x − 2) Include 0 x w hen you write out f( x). HintExercise 7B
|
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] |
142
Chapter 7
7 Show that
x3 + 2x2 − 5x − 10 = (x + 2)(x2 − 5)
8 Find the remainder when:
a x3 + 4x2 − 3x + 2 is divided by (x + 5) b 3x3 − 20x2 + 10x + 5 is divided by (x − 6)
c −2x3 + 3x2 + 12x + 20 is divided by (x − 4)
9 Show that w
hen 3x3 − 2x2 + 4 is divided by (x − 1) the remainder is 5.
10 Show that w
hen 3x4 − 8x3 + 10x2 − 3x − 25 is divided by (x + 1) the remainder is −1.
11 Show that (
x + 4) is a factor of 5x3 − 73x + 28.
12 Simplify 3x3 − 8x − 8 ___________ x − 2
13 Divide x3 − 1 by (x − 1).
14 Divide x4 − 16 by (x + 2).
15 f(x)
= 10x3 + 43x2 – 2x − 10
Find the remainder when f(x) is divided by (5x + 4). (2 marks)
16 f(x)
= 3x3 – 14x2 – 47x – 14
a Find the remainder when f(
x) is divided by (x − 3). (2 marks)
b Given tha
t (x + 2) is a factor of f(x), factorise f(x)
completely. (4 marks)
17 a Find the remainder when
x3 + 6x2 + 5x – 12 is divided by
i x −
2,
ii x +
3. (3 marks)
b Hence, or otherwise, find a
ll the solutions to the equation x3 + 6x2 + 5x – 12 = 0. (4 marks)
18 f(x)
= 2x3 + 3x2 – 8x + 3
a Show that f(
x) = (2x – 1)(ax2 + bx + c) where a, b and c are constants to be found. (2 marks)
b Hence factorise f(x) complete
ly. (4 marks)
c Write down a
ll the real roots of the equation f(x) = 0. (2 marks)
19 f(x)
= 12x3 + 5x2 + 2x – 1
a Show that (4
x – 1) is a factor of f(x) and write f(x) in the form (4x – 1)(ax2 + bx + c).
(6 marks)
b Hence, show tha
t the equation 12x3 + 5x2 + 2x – 1 = 0 has exactly 1 real solution.
(2 marks) Divide 3 x3 − 8x − 8 by ( x − 2). Hint
Write x3 − 1 as x3 + 0x2 + 0x − 1. Hint
E
E/P
Write f( x) in the form
(x + 2)( ax2 + bx + c) then
factorise the quadratic
factor.Problem-solving
E/P
E/P
E/P
|
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] |
143
Algebraic methods
7.3 The factor theorem
The factor theorem is a quick way of finding simple linear factors of a polynomial.
■ The factor theorem stat
es that if f( x) is a
polynomial then:
• If f( p) = 0, then ( x – p) is a factor of f( x).
• If (x – p) is a factor of f( x), then f( p) = 0.
You can use the factor theorem to quickly factorise a cubic function, g(x):
1 Substitut
e values into the function until you find a value p such that g(
p) =
0.
2 Divide the function by (x
– p).
3 Write g(
x) = (x – p)(ax2 + bx + c)
4 Factorise the quadr
atic factor, if possible, to write g(x) as a product of three linear factors.These two statements are not the same.
Here are two similar statements, only one of which is true:If x = – 2 then x
2 = 4 ✓
If x2 = 4 then x = – 2 ✗Watch out
The remainder will be 0 because (x – p) is a factor of g(x).
The other factor will be quadratic.
Example 5
Show that (x − 2) is a factor of x3 + x2 − 4x − 4 by:
a algebr
aic division b the factor theorem
a x2 + 3 x + 2
x 3 + x2 − 4 x − 4 ____ x − 2)
x3 − 2x2
3x2 − 4 x
3x2 − 6 x
2x − 4
2x − 4
0
So (x − 2) is a factor of x3 + x2 − 4 x − 4.
b f(x) = x3 + x2 − 4 x − 4
f(2) = (2)3 + (2)2 − 4(2) − 4
= 8 + 4 − 8 − 4 = 0
So (x − 2) is a factor of x
3 + x2 − 4 x − 4.Divide x3 + x2 − 4x − 4 by (x − 2).
The remainder is 0, so (x − 2) is a factor of
x3 + x2 − 4x − 4.
Write the polynomial as a function.
Substitute x = 2 into the polynomial.Use the factor theorem:
If f(
p) =
0, then (x − p) is a factor of f(x).
Here p = 2, so (x − 2) is a factor of x3 + x2 − 4x − 4.
|
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] |
144
Chapter 7
Example 6
a Fully factorise 2x3 + x2 – 18x – 9 b Hence sketch the gra ph of y = 2x3 + x2 − 18x − 9
a f(x) = 2 x3 + x2 − 18x − 9
f(−1) = 2(− 1)3 + (−1)2 − 18( −1) − 9 = 8
f(1) = 2(1)3 + (1)2 − 18(1) − 9 = − 24
f(2) = 2(2)3 + (2)2 − 18(2) − 9 = − 25
f(3) = 2(3)3 + (3)2 − 18(3) − 9 = 0
So (x − 3) is a factor of
2x3 + x2 − 18x − 9.
2x2 + 7x + 3
2 x 3 + x2 − 18x − 9 ____ x − 3)
2x3 − 6 x2
7x2 − 18x
7x2 − 21x
3x − 9
3x − 9
0
2x3 + x2 − 18x − 9 = (x − 3)(2 x2 + 7x + 3)
= (x − 3)(2 x + 1)( x + 3)
b 0 = (x − 3)(2 x + 1)( x + 3)
So the curve crosses the x -axis at (3, 0),
(− 1 __ 2 , 0) and ( −3, 0).
When x = 0, y = ( −3)(1)(3) = − 9
The curve crosses the y -axis at (0, − 9).
x → ∞ , y → ∞
x → − ∞, y → − ∞
–9
y = 2 x3 + x2 – /one.ss018x – 9–31
2–y
x OWrite the polynomial as a function.
Try values of x, e.g. −1, 1, 2, 3, … until you find
f( p) =
0.
f( p) = 0.
Use statement 1 from the factor theorem:If f(
p) =
0, then (x − p) is a factor of f(x).
Here p = 3.
Use long division to find the quotient when dividing by (x − 3).
You can check your division here:(x − 3) is a factor of 2x
3 + x2 − 18x − 9, so the
remainder must be 0.
2x2 + 7x + 3 can also be factorised.
Set y = 0 to find the points where the curve crosses the x-axis.
This is a cubic graph with a positive coefficient of x
3 and three distinct roots. You should be familiar
with its general shape. ← Section 4.1Set x = 0 to find the y-intercept.
|
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145
Algebraic methods
Example 7
Given that (x + 1) is a factor of 4x4 − 3x2 + a, find the value of a.
f(x) = 4 x 4 − 3 x2 + a
f(−1) = 0
4(−1)4 − 3(−1)2 + a = 0
4 − 3 + a = 0
a = −1Write the polynomial as a function.
Use statement 2 from the factor theorem.
(x − p) is a factor of f(x), so f( p) =
0
Here p = −1.
Substitute x = −1 and solve the equation for a.
Remember (−1)4 = 1.
1 Use the factor theorem to show tha t:
a (x
− 1) is a factor of 4x3 − 3x2 − 1 b (x + 3) is a factor of 5x 4 − 45x2 − 6x − 18
c (x
− 4) is a factor of −3x3 + 13x2 − 6x + 8.
2 Show that (
x − 1) is a factor of x3 + 6x2 + 5x − 12 and hence factorise the expression completely.
3 Show that (
x + 1) is a factor of x3 + 3x2 − 33x − 35 and hence factorise the expression completely.
4 Show that (
x − 5) is a factor of x3 − 7x2 + 2x + 40 and hence factorise the expression completely.
5 Show that (
x − 2) is a factor of 2x3 + 3x2 − 18x + 8 and hence factorise the expression completely.
6 Each of these expr
essions has a factor (x ± p). Find a value of p and hence factorise the
expression completely.
a x3 − 10x2 + 19x + 30 b x3 + x2 − 4x − 4 c x3 − 4x2 − 11x + 30
7 i Fully factorise the right-hand side of each equa
tion.
ii Sketch the gra
ph of each equation.
a y =
2x3 + 5x2 − 4x − 3 b y = 2x3 − 17x2 + 38x − 15 c y = 3x3 + 8x2 + 3x − 2
d y =
6x3 + 11x2 − 3x − 2 e y = 4x3 − 12x2 − 7x + 30
8 Given tha
t (x − 1) is a factor of 5x3 − 9x2 + 2x + a, find the value of a.
9 Given tha
t (x + 3) is a factor of 6x3 − bx2 + 18, find the value of b.
10 Given tha
t (x − 1) and (x + 1) are factors of px3 + qx2 − 3x − 7,
find the values of p and q.
11 Given tha
t (x + 1) and (x − 2) are factors of cx3 + dx2 – 9x – 10,
find the values of c and d.
12 Given tha
t (x + 2) and (x − 3) are factors of gx3 + hx2 – 14x + 24, find the values of g and h.P
P
P
Use the factor theorem
to form simultaneous equations.Problem-solving
P
PExercise 7C
|
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] |
146
Chapter 7
7.4 Mathematical proof
A proof is a logical and structured argument to show that a
mathematical statement (or conjecture) is always true. A mathematical proof usually starts with previously established mathematical facts (or theorems) and then works through a series of logical steps. The final step in a proof is a statement of what has been proven.
Known facts
or theoremsClearly shown
logical stepsStatement
of proo f
A mathematical proof needs to show that something is true in every case.
■ You can pro
ve a mathematical statement is true by deduction. This means starting from
known facts or definitions, then using logical steps to reach the desired conclusion.
Here is an example of proof by deduction:
Statement: The product of two odd numbers is odd.Demonstration: 5 × 7 = 35, which is odd
Proof: p and q are integers, so 2p + 1 and 2q + 1
are odd numbers.
(2p + 1) × (2q + 1)
= 4pq
+ 2p + 2q + 1
= 2(2pq + p + q) + 1
Since p and q are integers, 2pq + p + q is also an integer.
So 2(2pq + p + q) + 1 is one more than an even number.So the product of two odd numbers is an odd number. A stat ement that has
been proven is called a theorem .
A statement that has yet to be
proven is called a conjecture .Notation
This is demonstration but it is not a proof. You have only shown one case.
You can use 2p + 1 and 2q + 1 to represent any odd numbers. If you can show that (2p + 1) × (2q + 1) is always an odd number then you have proved the statement for all cases.
This is the statement of proof.13 f(x) = 3x3 – 12x2 + 6x − 24
a Use the factor theorem to show tha
t (x − 4) is a factor of f(x). (2 marks)
b Hence, show tha
t 4 is the only real root of the equation f(x) = 0. (4 marks)
14 f(x
) = 4x3 + 4x2 − 11x − 6
a Use the factor theorem to show tha
t (x + 2) is a factor of f(x). (2 marks)
b Factorise f(x
) completely. (4 marks)
c Write down a
ll the solutions of the equation 4x3 + 4x2 − 11x − 6 = 0. (1 mark)
15 a Show that (
x − 2) is a factor of 9x4 – 18x3 – x2 + 2x. (2 marks)
b Hence, find four r
eal solutions to the equation 9x 4 − 18x3 − x2 + 2x = 0. (5 marks)E
E
E
f(x) = 2 x4 − 5x3 − 42 x2 − 9x + 54
a Sho
w that f(1) = 0 and f( −3) = 0.
b Hen
ce, solve f( x) = 0.Challenge
|
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] |
147
Algebraic methods
Example 8
Prove that (3x + 2)(x − 5)(x + 7) ≡ 3x3 + 8x2 − 101x − 70
(3x + 2)( x – 5)( x + 7)
= (3x + 2)( x2 + 2x – 35)
= 3x3 + 6 x2 – 105 x + 2x2 + 4 x – 70
= 3x3 + 8 x2 – 101 x – 70
So
(3x + 2)(x – 5)(x + 7) ≡ 3 x3 + 8x2 – 101x – 70
Example 9
Prove that if (x − p) is a factor of f(x) then f( p) = 0.
If (x − p) is a factor of f( x) then
f(x) = (x − p) × g( x)
So f( p) =
( p
− p) × g( p)
i.
e. f( p) =
0 × g( p)
So f
(
p) = 0 as r
equired.g(x) is a polynomial expression.
Substitute x = p.
p − p = 0
Remember 0 × anything = 0Start with the left-hand side and expand the
brackets.
In proof questions you need to show all your working.
Left-hand side = right-hand side.■ In a mathematical proof y ou must
• State any information or assumptions you are using
• Show every step of your proof clearly
• Make sure that every step follows logically from the previous step
• Make sure you have covered all possible cases
• Write a statement of proof at the end of your working
You need to be able to prove results involving
identities, such as (a + b)(a – b) ≡ a2 – b2
■ To prove an identity you should• Star
t with the expression on one side
of the identity
• Manipulate that expression algebraically until it matches the other side
• Show every step of your algebraic working The s ymbol ≡ means ‘is
always equal to’. It shows that two
expressions are mathematically identical .Notation
Don ’t try to ‘solve’ an identity
like an equation. Start from one side and manipulate the expression to match the other side.Watch out
|
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] |
148
Chapter 7
Example 11
The equation kx2 + 3kx + 2 = 0, where k is a constant, has no real roots.
Prove that k satisfies the inequality 0 < k < 8 _ 9
kx2 + 3 kx + 2 = 0 has no real roots,
so b2 – 4 ac < 0
(3k)2 – 4k(2) < 0
9k2 – 8k < 0
k(9k – 8) < 0
y
k Oy = k(9k – 8)
8
9State which assumption or information you are
using at each stage of your proof.
The graph shows that when k(9k – 8) < 0, 0 < k <
8 _ 9 Use the discriminant. ← Sec tion 2.5
Solve this quadratic inequality by sketching the
graph of y = k (9k – 8) ← Sec tion 3.5Example 10
Prove that A(1, 1), B(3, 3) and C (4, 2) are the vertices of a right-angled triangle.
y
x OCB
A
The gradient of line AB = 3 − 1 ______ 3 − 1 = 2 __ 2 = 1
T
he gradient of line BC = 2 − 3 ______ 4 − 3 = −1 ___ 1 = −1
Th
e gradient of line AC = 2 − 1 ______ 4 − 1 = 1 __ 3
The g
radients are different so the three
points are not collinear.
Hence ABC is a triangle.
Gradient of AB × gradient of BC = 1 ×
(−1)
= −1
So AB is perpendicular to BC ,
and the triangle is a right-angled triangle.If you need to prove a geometrical
result, it can sometimes help to sketch a diagram as part of your working.Problem-solving
The gradient of a line = y2 − y1 ______ x2 − x1
If the product of two gradients is − 1 then the two
lines are perpendicular.
Gradient of line AB × gradient of line BC = −1
Remember to state what you have proved.
|
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] |
149
Algebraic methods
1 Prov
e that n2 − n is an even number for all values of n.
2 Prov
e that x ______ 1 + √ __
2 ≡ x √ __
2 − x.
3 Prov
e that (x + √ __
y )(x − √ __
y ) ≡ x2 − y.
4 Prov
e that (2x − 1)(x + 6)(x − 5) ≡ 2x3 + x2 − 61x + 30.
5 Prov
e that x2 + bx ≡ (x + b __ 2 ) 2
− ( b __ 2 ) 2
6 Prov
e that the solutions of x2 + 2bx + c = 0 are x = −b ± √ _____ b2 − c .
7 Prov
e that (x − 2 __ x ) 3
≡ x3 − 6x + 12 ___ x − 8 __ x3
8 Prov
e that (x3 − 1 __ x ) ( x 3 _ 2 + x − 5 _ 2 ) ≡ x 1 _ 2 (x 4 − 1 __ x4 )
9 Use completing the square to pro
ve that 3n2 − 4n + 10 is
positive f
or all values of n.
10 Use completing the square to pro
ve that −n2 − 2n − 3 is
negative for all values of n.
11 Prov
e that x2 + 8x + 20 > 4 for all values of x. (3 marks)
12 The equation kx2 + 5kx + 3 = 0, where k is a constant, has no real roots. Prove that k satisfies
the inequality 0 < k , 12 ___ 25 (4 marks)P The proofs in this
exercise are all proofs by
deduction.Hint
P
P
P
P
P
P
P
P
Any expression that is squared must be > 0.Problem-solving
P
E/P
E/PExercise 7D0 < k < 8 __ 9
Whe
n k = 0:
(0)x2 + 3(0) x + 2 = 0
2
= 0
Which is impossible, so no real roots
So combining these:
0 < k < 8 __ 9 as requiredBe really careful to consider all the possible
situations. You can’t use the discriminant if k = 0 so look at this case separately.
Write out all of your conclusions clearly.
0 < k < 8 _ 9 together with k = 0, gives 0 < k < 8 _ 9
|
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] |
150
Chapter 7
7.5 Methods of proo f
A mathematical statement can be proved by exhaustion. For example, you can prove that the sum
of two consecutive square numbers between 100 and 200 is an odd number. The square numbers between 100 and 200 are 121, 144, 169, 196.
121 + 144 = 265 which is odd
144 + 169
= 313 which is odd 169 + 196
= 365 which is odd
So the sum of two consecutive square numbers between 100 and 200 is an odd number.■
You can pro
ve a mathematical statement is true by exhaustion. This means breaking the
statement into smaller cases and proving each case separately.
This method is better suited to a small number of results. You cannot use one example to prove a
statement is true, as one example is only one case.13 The equation px2 – 5x − 6 = 0, where p is a constant, has two distinct real roots. Prove that p
satisfies the inequality p > – 25 ___ 24 (4 marks)
14 Prov
e that A(3, 1), B(1, 2) and C (2, 4) are the vertices of
a right-angled triangle.
15 Prov
e that quadrilateral A(1, 1), B(2, 4), C(6, 5) and D(5, 2) is a parallelogram.
16 Prov
e that quadrilateral A(2, 1), B(5, 2), C (4, −1) and
D(1, −2) is a rhombus.
17 Prov
e that A(−5, 2), B(−3, −4) and C (3, −2) ar
e the vertices of an isosceles right-angled triangle.
18 A circle has equation (
x − 1)2 + y2 = k, where k > 0.
The straight line L with equation y = ax cuts the circle at two distinct points.
Prove that k > a2 ______ 1 + a2 (6 marks)
19 Prov
e that the line 4y − 3x + 26 = 0 is a tangent to the circle (x + 4)2 + ( y − 3)2 = 100. (5 marks)
20 The diagram sho
ws a square and four
congruent right-angled triangles.
Use the diagram to prove that a2 + b2 = c2.E/P
P
P
P
P
E/P
E/P
P
cb
aFind an expression for the
area of the large square in terms of a and b .Problem-solving
1 Prove that A (7 , 8), B (−1, 8), C (6, 1 ) and D (0, 9) are points on the same circle.
2 Pro
ve that any odd prime number can be written as the difference of two squares.Challenge
|
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] |
151
Algebraic methods
A mathematical statement can be disproved using a counter-example. For example, to prove that
the statement ‘3n + 3 is a multiple of 6 for all values of n’ is not true you can use the counter-example when n = 2, as 3 × 2 + 3 = 9 and 9 is not a multiple of 6.
■
You can pro
ve a mathematical statement is not true by a counter-example. A counter-
example is one example that does not work for the statement. You do not need to give more
than one example, as one is sufficient to disprove a statement.Example 12
Prove that all square numbers are either a multiple of 4 or 1 more than a multiple of 4.
For odd numbers:
(2n + 1)2 = 4 n2 + 4 n + 1 = 4 n(n + 1) + 1
4n(n + 1) is a multiple of 4, so 4 n(n + 1) + 1 is
1 more than a multiple of 4.For even numbers:(2n)
2 = 4 n2
4n2 is a multiple of 4.
All integers are either odd or even, so all
square numbers are either a multiple of 4 or 1 more than a multiple of 4.You can write any odd number in the form 2n + 1
where n is a positive integer.
You can write any even number in the form 2n where n is a positive integer.
Example 13
Prove that the following statement is not true:
‘The sum of two consecutive prime numbers is always even.’
2 and 3 are both prime
2 + 3 = 55 is odd
So the statement is not true.You only need one counter-example to show that
the statement is false.Consider the two cases, odd and even numbers, separately.Problem-solving
|
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] |
152
Chapter 7
1 Prov
e that when n is an integer and 1 < n < 6, then m = n + 2 is
not divisible by 10.
2 Prov
e that every odd integer between 2 and 26 is either prime or the product of two primes.
3 Prov
e that the sum of two consecutive square numbers between 12 to 82 is an odd number.
4 Prov
e that all cube numbers are either a multiple of 9 or 1 more or 1 less than a multiple of 9.
(4 marks)P You can try each integer
for 1 < n < 6.Hint
P
P
E/PExercise 7EExample 14
a Prove that for all positive values of x and y:
x __ y + y
__ x > 2
b Use a counter-example to sho
w that this is not true when
x and y are not both positive. You m ust always
start a proof from known facts .
Never start your proof with the
statement you are trying to prove.Watch out
a Jottings:
x __ y + y __ x > 2
x2 + y2 _______ xy > 2
x2 + y2 − 2xy > 0
(x − y)2 > 0
Proof:
Consider ( x − y)2
(x − y)2 > 0
x2 + y2 − 2xy > 0
x2 + y2 + 2xy _____________ xy > 0
T
his step is valid because x and y are
both positive so xy > 0.
x __ y + y __ x − 2 > 0
x __ y + y __ x > 2
b Try
x = −3 and y = 6
−3 ___ 6 + 6 ___ −3 = − 1 __ 2 − 2 = − 5 __ 2
Thi
s is not > 2 so the statement is not
true.Use jottings to get some ideas for a good starting
point. These don’t form part of your proof, but can give you a clue as to what expression you can consider to begin your proof.Problem-solving
Now you are ready to start your proof. You know that any expression squared is > 0. This is a known
fact so this is a valid way to begin your proof.
State how you have used the fact that x and y are positive in your proof. If xy = 0 you couldn’t divide the RHS by xy, and if xy < 0, then the direction of the inequality would be reversed.
This was what you wanted to prove so you have finished.
Your working for part a tells you that the proof fails when xy < 0, so try one positive and one negative value.
|
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] |
153
Algebraic methods
5 Find a counter-example to dispr
ove each of the following statements:
a If n
is a positive integer then n4 − n is divisible by 4.
b Integers alw
ays have an even number of factors.
c 2n2 − 6n + 1 is positive for all values of n.
d 2n2 − 2n − 4 is a multiple of 3 for all integer values of n.
6 A student is trying to prov
e that x3 + y3 < (x + y)3.
The student writes:
(x + y)3 = x3 + 3 x2y + 3 xy2 + y3
which is less than x3 + y3 since
3x2y + 3 xy2 > 0
a Identify the error made in the proof . (1 mark)
b Provide a counter
-example to show that the statement is not true. (2 marks)
7 Prov
e that for all real values of x
(x + 6)2 > 2x + 11 (3 marks)
8 Given tha
t a is a positive real number, prove that:
a + 1 __ a > 2
(2 marks)
9 a Pro
ve that for any positive numbers p and q:
p + q > √ ____ 4pq (3 marks)
b Show, b
y means of a counter-example, that this
inequality does not hold when p and q are both
negative. (2 marks)
10 It is claimed that the follo
wing inequality is true for all negative numbers x and y:
x + y > √ ______ x2 + y2
The following pr
oof is offered by a student:
x + y > √ _______ x2 + y2
(x +
y)2 > x2 + y2
x2 + y2 + 2xy > x2 + y2
2xy > 0 which is true because x and
y are both negative, so xy is positive.
a Explain the error made by the student. (2 marks)
b By use of a counter-e
xample, verify that the inequality is not satisfied if both
x and y are negative. (1 mark)
c Prov
e that this inequality is true if x and y are both positive. (2 marks)P
E/P
For part b you need to write
down suitable values of x and y and show that they do not satisfy the inequality.Problem-solving
E/P
E/P
Rem ember to state how
you use the condition that a is positive.Watch out
E/P
Use jottings and work backwards to work out what expression to consider.Problem-solving
E/P
|
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] |
154
Chapter 7
1 Simplify these fractions as far as possible:
a 3x 4 − 21x _________ 3x
b x2 − 2x − 24 ___________ x2 − 7x + 6
c 2x2 + 7x − 4 ___________ 2x2 + 9x + 4
2 Divide 3x3 + 12x2 + 5x + 20 by (x + 4).
3 Simplify 2x3 + 3x + 5 ___________ x + 1
4 a Show that (
x − 3) is a factor of 2x3 − 2x2 − 17x + 15. (2 marks)
b Hence express 2
x3 − 2x2 − 17x + 15 in the form (x − 3)(Ax2 + Bx + C ), where the v alues
A, B and C are to be found. (3 marks)
5 a Show that (
x − 2) is a factor of x3 + 4x2 − 3x − 18. (2 marks)
b Hence express
x3 + 4x2 − 3x − 18 in the form (x − 2)( px + q)2, where the values p and q
are to be found. (4 marks)
6 Factorise completel
y 2x3 + 3x2 − 18x + 8. (6 marks)
7 Find the value of
k if (x − 2) is a factor of x3 − 3x2 + kx − 10. (4 marks)
8 f(x) = 2x2 + px + q. Given that f(− 3) = 0, and f(4) = 21:
a find the value of
p and q (6 marks)
b factorise f(x ). (3 marks)
9 h(x) = x3 + 4x2 + rx + s. Given h(− 1) = 0, and h(2) = 30:
a find the values of
r and s (6 marks)
b factorise h(x ). (3 marks)
10 g(x) = 2x3 + 9x2 − 6x − 5.
a Factorise g(x ). (6 marks)
b Solve g(x ) = 0. (2 marks)E
E
E
E/P
E/P
E/P
EMixed exercise 7
|
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] |
155
Algebraic methods
11 a Show that (
x − 2) is a factor of f(x ) = x3 + x2 − 5x − 2. (2 marks)
b Hence, or otherwise, find the e
xact solutions of the equation f(x ) = 0. (4 marks)
12 Given tha
t −1 is a root of the equation 2x3 − 5x2 − 4x + 3, find the two positive roots. (4 marks)
13 f(x
) = x3 – 2x2 – 19x + 20
a Show that (
x + 4) is a factor of f(x). (3 marks)
b Hence, or otherwise, find a
ll the solutions to the equation
x3 – 2x2 – 19x + 20 = 0. (4 marks)
14 f(x
) = 6x3 + 17x2 – 5x − 6
a Show that f(
x) = (3x – 2)(ax2 + bx + c), where a , b and c are constants to be found. (2 marks)
b Hence factorise f(x) complete
ly. (4 marks)
c Write down a
ll the real roots of the equation f(x) = 0. (2 marks)
15 Prov
e that x −
y _______ √ __
x − √ __
y ≡ √ __
x + √ __
y .
16 Use completing the square to pro
ve that n2 − 8n + 20 is positive for all values of n.
17 Prov
e that the quadrilateral A(1, 1), B(3, 2), C (4, 0) and D(2,
−1) is a square.
18 Prov
e that the sum of two consecutive positive odd numbers less than ten gives an even
number.
19 Prov
e that the statement ‘n2 − n + 3 is a prime number for all values of n’ is untrue.
20 Prov
e that (x − 1 __ x ) ( x 4 _ 3 + x − 2 _ 3 ) ≡ x 1 _ 3 (x2 − 1 __ x2 ) .
21 Prov
e that 2x3 + x2 − 43x – 60 ≡ (x + 4)(x – 5)(2x + 3).
22 The equation x2 – kx + k = 0, where k is a positive constant, has two equal roots.
Prove that k = 4. (3 marks)
23 Prov
e that the distance between opposite edges of a regular hexagon of side length √ __
3 is a
rational value.E
E
E
E
P
P
P
P
P
P
E
P
|
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] |
156
Chapter 7
24 a Prov
e that the difference of the squares of two consecutive even numbers is always divisible
by 4.
b Is this statement true for od
d numbers? Give a reason for your answer.
25 A student is trying to prov
e that 1 + x2 < (1 + x)2.
The student writes:
(1 + x)2 = 1 + 2x + x2.
So 1 + x2 < 1 + 2x + x2.
a Identify the error made in the proof . (1 mark)
b Provide a counter
-example to show that the statement is not true. (2 marks)P
E
1 The diagram shows two squares and a circle.
a Given that π is defined as the circumference of a circle of diameter 1 unit,
prove
that 2 √ __
2 < π < 4.
b By si
milarly constructing regular hexagons inside and outside a circle,
prove that 3 < π < 2 √ __
3 .
2 Prove t
hat if f( x) = ax3 + bx2 + cx + d and f( p) = 0, then ( x − p ) is a factor of f( x).Challenge
|
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] |
157
Algebraic methods
1 When simplifying an algebr aic fraction, factorise the numerator and denominator where
possible and then cancel common factors.
2 You can use long division t
o divide a polynomial by (x ± p), where p is a constant.
3 The factor theor
em states that if f(x) is a polynomial then:
• If f( p) = 0, then (x
– p) is a factor of f(x)
• If (x
– p) is a factor of f(x), then f( p) =
0
4 You can pro
ve a mathematical statement is true by deduction. This means starting from
known factors or definitions, then using logical steps to reach the desired conclusion.
5 In a mathematical proof y
ou must
• State any information or assumptions you are using
• Sho
w ever
y step of your proo
f clearly
•
Make sure that ever
y step foll
ows logically from the previous step
•
Make sure you hav
e co
vered all possibl
e cases
•
Write a statement of proo
f at the end of your working
6 To pro
ve an identity you should
• Start with the expr
ession on one side of the identity
•
Manipulate that expr
ession algebr
aically until it matches the other side
• Sho
w ever
y step of your algebr
aic working
7 You can pro
ve a mathematical statement is true by exhaustion. This means breaking the
statement into smaller cases and proving each case separately.
8 You can pro
ve a mathematical statement is not true by a counter-example. A counter-
example is one example that does not work for the statement. You do not need to give more
than one example, as one is sufficient to disprove a statement.Summary of key points
|
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] |
1588
The binomial
expansion
After completing this chapter you should be able to:
● Use Pascal
’s triangle to identify binomial coefficients and use them
to expand simple binomial expressions → pages 159–161
● Use combinations and factorial notation → pages 161–163
● Use the binomial expansion t o expand brackets → pages 163–165
● Find individual coefficients in a binomial exp ansion → pages 165–167
● Make approximations using the binomial expansion → pages 167–169Objectives
1 Expand and simplify where possible:
a (2x − 3
y)2 b (x − y)3 c (2 + x)3
← Section 1.2
2 Simplifya
(−2x)3 b (3x)−4
c ( 2 __ 5 x) 2
d ( 1 __ 3 x) −3
← Sections 1.1, 1.4
3 Simplify
a (25x
) 1 _ 2 b (64x) − 2 _ 3
c ( 9 ____ 100 x) − 1 _ 2
d ( 8 ___ 27 x) 4 _ 3
← Section 1.4Prior knowledge check
The binomial expansion can be used to
expand brackets raised to large powers. It can be used to simplify probability models with a large number of trials, such as those used by manufacturers to predict faults.
→ Exercise 8E Q9
|
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] |
159
The binomial expansion
8.1 Pascal’s triangle
You can use Pascal’s triangle to quickly expand expressions such as (x + 2y)3.
Consider the expansions of (a + b)n for n = 0, 1, 2, 3 and 4:
(a + b)0 =
(a + b)1 =
(a + b)2 =
(a + b)3 =
(a + b)4 = + 1a44a3b +6a2b2+4ab3+ 1a33a2b+ 3ab2+ 1a22ab+ 1a1
1b
+ 1b2
+ 1b3
+1b4
Every term in the expansion of (a + b)n has total index n:
In the 6a2b2 term the total index is 2 + 2 = 4.
In the 4ab3 term the total index is 1 + 3 = 4.
The coefficients in the expansions form a pattern that is known as Pascal’s triangle.
■ Pascal’s triangle is formed by adding adjacent pairs of numbers to find the numbers on the
next r
ow.
Here are the first 7 rows of Pascal’s triangle:
+
+
16 15 20 15 615 10 10 511464 11331121111
1
■ The ( n + 1)th row of Pascal’s triangle gives the coefficients in the expansion of ( a + b)n.Each coefficient is the sum of the two
coefficients immediately above it.
The third row of Pascal’s triangle gives the coefficients in the expansion of
(a + b) 2 .
a (x + 2y)3
The coefficients are 1, 3, 3, 1 so:
(x + 2y)3 = 1x3 + 3x2(2y) + 3x(2y)2 + 1(2y)3
= x3 + 6 x2y + 12 xy2 + 8 y3Example 1
Index = 3 so look at the 4th row of Pascal’s
triangle to find the coefficients.Use Pascal’s triangle to find the expansions of:
a (x
+ 2y)3 b (2x − 5)4
This is the expansion of (a + b)3 with a = x and
b = 2y. Use brackets to make sure you don’t make
a mistake. (2y)2 = 4y2.
|
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] |
160
Chapter 8
The coefficients are 1, 3, 3, 1:
The term in x2 is 3 × 2(− cx)2 = 6 c2x2
So 6 c2 = 294
c2 = 49
c = ±7Example 2Index = 4 so look at the 5th row of Pascal’s
triangle.
Index = 3 so use the 4th row of Pascal’s triangle.The coefficient of x2 in the expansion of (2 − cx)3 is 294.
Find the possible value(s) of the constant c.b (2x − 5)4
The coefficients are 1, 4, 6, 4, 1 so:
(2x − 5)4 = 1(2x)4 + 4(2 x)3(−5)1
+ 6(2 x)2(−5)2 + 4(2 x)1(−5)3
+ 1(−5)4
= 16x4 − 160 x3 + 600 x2
− 1000 x + 625Be careful with the negative numbers.
1 State the row of Pascal’s triangle that would give the coefficients of each expansion:
a (x
+ y)3 b (3x − 7)15 c (2x + 1 _ 2 ) n d (y − 2 x)n + 4
2 Write down the expansion of:
a (x
+ y)4 b (p + q)5 c (a − b)3 d (x + 4)3
e (2x − 3)4 f (a + 2)5 g (3x − 4)4 h (2x − 3y)4
3 Find the coefficient of x3 in the expansion of:
a (4 +
x)4 b (1 − x)5 c (3 + 2 x)3 d (4 + 2 x)5
e (2 + x)6 f (4 − 1 _ 2 x) 4 g (x + 2)5 h (3 − 2x)4
4 Fully expand the expression (1 + 3x)(1 + 2x)3.
5 Expand (2 + y
)3. Hence or otherwise, write down the expansion of (2 + x − x2)3 in ascending
powers of x.
6 The coefficient of x2 in the expansion of (2 + ax)3 is 54. Find the possible values of the constant a .P
P
PExercise 8AIf there is an unknown in the original
expression, you might be able to form an equation involving that unknown.Problem-solving
Expand (1 + 2x)3, then multiply each
term by 1 and by 3 x.Problem-solvingThis is the expansion of (a + b)4 with a = 2x and
b = −5.
From the expansion of (a + b)3 the x2 term is 3ab2
where a = 2 and b = –cx.
Form and solve an equation in c.
|
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] |
161
The binomial expansion
7 The coefficient of x3 in the expansion of (2 − x)(3 + bx)3 is 45. Find possible values of the
constant b.
8 Work out the coefficient of
x2 in the expansion of ( p − 2x)3. Give your answer in terms of p.
9 After 5 years, the v
alue of an investment of £500 at an interest rate of X % per annum is giv
en by:
500 (1 + X ____ 100 ) 5
Find an approximation for this expression in the form A + BX + CX 2, where A, B and C are
constants to be found. Y ou can ignore higher powers of X.P
P
P
8.2 Factorial notation
You can use combinations and factorial notation to help you expand binomial expressions. For larger
indices, it is quicker than using Pascal’s triangle.
Using factorial notation 3 × 2 × 1 = 3!
■ You can use factorial notation and your calculator to find entries in Pascal’s triangle quickly.
• The number of w
ays of choosing r items from a group of
n items is written as nCr or ( n r ) :
nCr = ( n r ) = n ! ________ r !(n − r)!
• The r th entry in the nth row of Pascal’s triangle is
given by n − 1Cr − 1 = ( n − 1 r − 1 ) You s ay ‘n factorial’.
By definition, 0! = 1.Notation
You c an say
‘n choose r ’ for nCr . It is sometimes
wr
itten without superscripts and
subscripts as nCr .NotationFind the constant term in the expansion of (x2 − 1 ___ 2x ) 3
.Challenge
a 5! = 5 × 4 × 3 × 2 × 1 = 120
b 5C2 = 5 ! ____ 2 !3 ! = 120 ____ 12 = 10
c 9C5 = 126Example 3
Calculate:
a 5! b 5C2 c the 6th entry in the 10th row of Pascal’s triangle
The rth entry in the n th row is n − 1Cr − 1. You can calculate 5C2 by using the nCr function on
your calculator.
nCr = n ! _______ r ! (n − r) ! = 5 ! ________ 2 ! (5 − 2) !
In the expansion of (a + b)9 this would give the
term 126a4b5. Use the nCr and ! functions on
your calculator to answer this question.Online
|
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] |
162
Chapter 8
1 Work out:
a 4! b 9! c 10 ! ___ 7 ! d 15 ! ___ 13 !
2 Without using a calcula
tor, work out:
a ( 4 2 ) b ( 6 4 ) c 6C3 d ( 5 4 ) e 10C8 f ( 9 5 )
3 Use a calculator to w
ork out:
a ( 15 6 ) b 10C7 c ( 20 10 ) d ( 20 17 ) e 14C9 f 18C5
4 Write each value a to d from
Pasca
l’s triangle using
nCr notation:
5 Work out the 5th n
umber on the 12th row from Pascal’s triangle.
6 The 11th row of
Pascal’s triangle is shown below.
1 10 45 … …
a Find the next two v
alues in the row.
b Hence find the coefficient of x3 in the expansion of (1 + 2x)10.
7 The 14th row of
Pascal’s triangle is shown below.
1 13 78 … …
a Find the next two v
alues in the row.
b Hence find the coefficient of x4 in the expansion of (1 + 3x)13.
8 The probability of
throwing exactly 10 heads when a fair coin is tossed 20 times is given by
( 20 10 ) 0. 5 20 . Calculate the probability and describe the likelihood of this occurring.
9 Show that:
a nC1 = n b n C 2 = n(n − 1) _______ 2
10 Given tha
t ( 50 13 ) = 50 ! _____ 13 !a ! , write down the value of a. (1 mark)
11 Given tha
t ( 35 p ) = 35 ! _____ p !18 ! , write down the value of p. (1 mark)16 cd 15 615 b 10 511 a 64 11331121111
1
P
E
EExercise 8B
|
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] |
163
The binomial expansion
8.3 The binomial expansion
The binomial expansion is a rule that allows you to expand brackets. You can use ( n r ) to work out the
coefficients in the binomial expansion. For example,
in the expansion of (a + b)5 = (a + b)(a + b)(a + b)(a + b)(a + b), to find the b3 term you can choose
multiples of b from 3 different brackets. You can do this in ( 5 3 ) ways so the b3 term is ( 5 3 ) a2b3.
■ The binomial expansion is:
(a + b) n = a n + ( n 1 ) a n − 1 b + ( n 2 ) a n − 2 b 2 + . . . + ( n r ) a n − r b r + . . . + b n (n ∈ ℕ)
where ( n r ) = n C r = n ! ________ r !(n − r) ! n ∈ ℕ mean s that n must be a
member of the natural numbers . This is all
the positive integers .Notation
(3 − 2x)5 = 35 + ( 5 1 ) 34(−2x) + ( 5 2 ) 33(−2x)2
+ ( 5 3 ) 32(−2x)3 + ( 5 4 ) 31(−2x)4
+ (−2x)5
= 243 − 810 x + 1080x2
− 720 x3 + 240 x4 − 32 x5Example 4
There are ( 5 2 ) ways of choosing two ‘−2x’ terms
from five brackets.There will be 6 terms.
Each term has a total index of 5.Use (a + b)
n where a = 3, b = −2x and n = 5.Use the binomial theorem to find the expansion of (3 − 2x)5.
Example 5
Find the first four terms in the binomial expansion of:
a (1 + 2
x)10 b (10 − 1 _ 2 x) 6 a Work out 10C3 and 10C7
b Wor
k out 14C5 and 14C9
c What do you notice about your answers to parts a and b ?
d Prove t
hat nCr = nCn−rChallenge
Work out each coefficient quickly
us
ing the nCr and power functions on your
calculator.Online
|
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] |
164
Chapter 8
a (1 + 2x)10
= 1 10 + ( 10 1 ) 1 9 (2x) + ( 10 2 ) 1 8 (2x) 2
+ ( 10 3 ) 1 7 (2x) 3 + . . .
= 1 + 20 x
+ 180 x2 + 960 x3 + . . .
b (10 − 1 _ 2 x) 6
= 10 6 + ( 6 1 ) 10 5 (− 1 _ 2 x) + ( 6 2 ) 10 4 (− 1 _ 2 x) 2
+ ( 6 3 ) 10 3 (− 1 _ 2 x) 3 + . . .
= 1 000 000 − 30
0 000 x +
37 500 x2
− 2500 x3 + . . . Write each coefficient in its simplest form. Thi s is sometimes called the
expansion in ascending powers of x .Notation
1 Write down the e
xpansion of the following:
a (1 +
x)4 b (3 + x)4 c (4 − x)4 d (x + 2)6 e (1 + 2 x)4 f (1 − 1 _ 2 x) 4
2 Use the binomial theorem to find the first f our terms in the expansion of:
a (1 +
x)10 b (1 − 2 x)5 c (1 + 3 x)6 d (2 − x)8 e (2 − 1 _ 2 x) 10 f (3 − x)7
3 Use the binomial theorem to find the first f our terms in the expansion of:
a (2x
+ y)6 b (2x + 3y)5 c (p − q)8 d (3x − y)6 e (x + 2y)8 f (2x − 3y)9
4 Use the binomial expansion to find the first f our terms, in ascending powers of x, of:
a (1 + x
)8 b (1 − 2 x)6 c (1 + x __ 2 ) 10
d (1 − 3 x)5 e (2 + x )7 f (3 − 2 x)3
g (2 − 3x )6 h (4 + x )4 i (2 + 5 x)7
5 Find the first 3 terms, in ascending po wers of x, of the binomial expansion of (2 − x)6
and simplify each term. (4 marks)
6 Find the first 3 terms, in ascending po
wers of x, of the binomial expansion of (3 − 2x)5
giving each term in its simplest form. (4 marks)
7 Find the binomial expansion of
(x + 1 __ x ) 5
giving each term in its simplest form. (4 marks) Your answers should be
in th
e form a + bx + cx2 + dx3
where a , b, c and d are numbers.Hint
E
E
E/PExercise 8C
a Show that ( a + b)4 − (a − b )4 = 8ab(a2 + b2).
b Giv
en that 82 896 = 174 − 54, write 82 896 as a product of its prime factors.Challenge
|
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] |
165
The binomial expansion
8.4 Solving binomial problems
You can use the general term of the binomial expansion to find individual coefficients in a binomial
expansion.
■ In the expansion of ( a + b)n the general term is given by ( n r ) an − rbr.
x3 term = ( 10 3 ) 17(kx)3 = 15 x3
120k3x3 = 15 x3
k = 1 __ 2 Example 7
a = 1, b = kx, n = 10 and r = 3.
k3x3 = 15 ____ 120 x3
k3x3 = 1 __ 8 x3
k3 = 1 __ 8 , k = 3 √ __
1 __ 8 g(x) = (1 + kx)10, where k is a constant.
Given that the coefficient of x3 in the binomial expansion of g(x) is 15, find the value of k.a x4 term = ( 10 4 ) 26(3x)4
= 210 × 64 × 81 x4
= 1 088 640 x4
The coefficient of x4 in the binomial expansion
of (2 + 3 x)10 is 1 088 640.
b (3 −
2x)7
= 37 + ( 7 1 ) 36(−2x) + ( 7 2 ) 35(−2x)2
+ ( 7 3 ) 34(−2x)3 + …
= 2187 − 10 206 x
+ 20 412x2
− 22 680 x3 + …
(2 + x )(2187 − 10 206 x
+ 20 412x2
− 22 680 x3 + …)
x3 term = 2 × (−22 680 x)3 + x × 20 412x2
= −24 948 x3
The coefficient of x3 in the binomial
expansion of (2 + x)(3 − 2x)7 is −24 948.Example 6
Use the general term. The power is 10, so n = 10,
and you need to find the x4 term so r = 4.
There are ( 10 4 ) ways of choosing 4 ‘3x’ terms
from 10 brackets.
First find the first four terms of the binomial
expansion of (3 − 2x)7.
Now expand the brackets (2 + x)(3 − 2x)7.a Find the coefficient of x4 in the binomial expansion of (2 + 3x)10.
b Find the coefficient of x3 in the binomial expansion of (2 + x)(3 − 2x)7.
There are two ways of making the x3 term:
(constant term × x3 term) and (x term × x2 term).
|
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] |
166
Chapter 8
a (1 + qx)8
= 18 + ( 8 1 ) 17(qx)1 + ( 8 2 ) 16(qx)2 + …
= 1 + 8 qx + 28 q2x2 + …
b 8q
= −r and 28 q2 = 7r
8q = −4 q2
4q2 + 8 q = 0
4q(q + 2) = 0
q = −2, r = 16Example 8
Using 28q2 = 7r, r = 4q2 and −r = −4q2.
q is non-zero so q = −2.a Write down the first three terms, in ascending powers of x, of the binomial expansion of
(1 +
qx)8, where q is a non-zero constant.
b Given tha
t, in the expansion of (1 + qx)8, the coefficient of x is −r and the coefficient of x2 is 7r,
find the value of q and the value of r.
1 Find the coefficient of x3 in the binomial expansion of:
a (3 +
x)5 b (1 + 2 x)5 c (1 − x)6 d (3x + 2)5
e (1 + x)10 f (3 − 2x)6 g (1 + x )20 h (4 − 3 x)7
i (1 − 1 _ 2 x) 6 j (3 + 1 _ 2 x) 7 k (2 − 1 _ 2 x) 8 l (5 + 1 _ 4 x) 5
2 The coefficient of x2 in the expansion of
(2 + ax)6 is 60. Find two possible values
of the constant a.
3 The coefficient of x3 in the expansion of (3 + bx)5 is −720. Find the value of the constant b.
4 The coefficient of x3 in the expansion of (2 + x)(3 − ax)4 is 30. Find two possible values of the
constant a.
5 When (1 −
2x)p is expanded, the coefficient of x2 is 40. Given that p > 0, use this information
to find:
a the value of
the constant p (6 marks)
b the coefficient of x (1 mark)
c the coefficient of x3 (2 marks)
6 a Find the first three terms
, in ascending powers of x, of the binomial expansion
of (5 + px)30, where p is a non-zero constant. (2 marks)
b Given tha
t in this expansion the coefficient of x2 is 29 times the coefficient of x
work out the value of p. (4 marks)P
P
P
E/P
E/PExercise 8DThere are two unknowns in this expression. Your
expansion will be in terms of q and x .Problem-solving
a = 2, b = ax , n = 6. Use brackets when you
substitute ax . Problem-solving
You will need to use the definition of ( n r ) to find
an expression for ( p 2 ) .Problem-solving
|
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] |
167
The binomial expansion
7 a Find the first four terms
, in ascending powers of x, of the binomial expansion
of (1 + qx)10, where q is a non-zero constant. (2 marks)
b Given tha
t in the expansion of (1 + qx)10 the coefficient of x3 is 108 times the
coefficient of x, work out the value of q. (4 marks)
8 a Find the first three terms
, in ascending powers of x of the binomial expansion
of (1 + px)11, where p is a constant. (2 marks)
b The first 3 terms in the same expansion ar
e 1, 77x and qx2, where q is a constant.
Find the value of p and the value of q. (4 marks)
9 a Write down the first thr
ee terms, in ascending powers of x, of the binomial
expansion of (1 + px)15, where p is a non-zero constant. (2 marks)
b Given tha
t, in the expansion of (1 + px)15, the coefficient of x is (−q) and the
coefficient of x2 is 5q, find the value of p and the value of q. (4 marks)
10 In the binomial expansion of
(1 + x)30, the coefficients of x9 and x10 are p and q respectively.
Find the value of q
__ p . (4 marks)E/P
E/P
E/P
E/P
8.5 Binomial estimation
In engineering and science, it is often useful to find simple approximations for complicated
functions. If the value of x is less than 1, then xn gets smaller as n gets larger. If x is small you can
sometimes ignore large powers of x to approximate a function or estimate a value.
Example 9
a Find the first four terms of the binomial expansion, in ascending powers of x, of (1 − x __ 4 ) 10
.
b Use your expansion to estima
te the value of 0.97510, giving your answer to 4 decimal places.Find the coefficient of x4 in the binomial expansion of: a (3 − 2 x2)9 b ( 5 __ x + x 2 ) 8
Challenge
a (1 − x __ 4 ) 10
= 1 10 + ( 10 1 ) 1 9 (− x __ 4 ) + ( 10 2 ) 1 8 (− x __ 4 ) 2
+ ( 10 3 ) 1 7 (− x __ 4 ) 3
+ . . .
= 1 −
2.5x + 2.8125 x2 − 1.875 x3 + …
|
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] |
168
Chapter 8
b We want (1 − x __ 4 ) = 0.975
x __ 4 = 0.025
x = 0.1
Substitute x = 0.1 into the expansion
for (1 − x __ 4 ) 10
from part a :
0.97510 ≈ 1 − 0.25 + 0.028 12 5
− 0.001 87
5
= 0.77625
0.97510 ≈ 0.7763 to 4 d.p.Substitute x = 0.1 into your expansion.
Using a calculator, 0.97510 = 0.776 329 62.
So approximation is c
orrect to 4 decimal places.
1 a Find the first four terms of the binomial expansion, in ascending powers of x, of (1 − x __ 10 ) 6
.
b By substituting an appropria
te value for x, find an approximate value for 0.996.
2 a Write do
wn the first four terms of the binomial expansion of (2 + x __ 5 ) 10
.
b By substituting an appropria
te value for x, find an approximate value for 2.110.
3 If x
is so small that terms of x3 and higher
can be ignored, show that:
(2 + x)(1 − 3x)5 ≈ 2 − 29x + 165x2
4 If x is so small tha t terms of x3 and higher
can be ignored, and
(2 − x)(3 + x)4 ≈ a + bx + cx2
find the values of the constants a, b and c.
5 a Write do
wn the first four terms in the expansion of (1 + 2x)8.
b By substituting an appropria
te value of x (which should be stated), find an approximate value
of 1.028.
6 f(x)
= (1 − 5x)30
a Find the first four terms , in ascending powers of x, in the binomial expansion of f(x).
b Use your answ
er to part a to estimate the value of (0.995)30, giving your answer to 6 decimal places.
c Use your calcula
tor to evaluate 0.99530 and calculate the percentage error in your answer to
part b.
7 a Find the first 3 terms
, in ascending powers of x , of the binomial expansion of (3 − x __ 5 ) 10
,
giving each term in its simplest form. (4 marks)
b Explain how you w
ould use your expansion to give an estimate for the value of 2.9810. (1 mark) Start by using the binomial expansion to
ex
pand (1 − 3 x)5. You can ignore terms of x3 and
higher so you only need to expand up to and
including the x2 term.Hint P
P
E/PExercise 8E
Find the first 3 terms in the expansion of (2 − x )(3 + x )
4, compare with a + bx + cx2 and
write down the values of a , b and c .Problem-solving Use GeoGebra to find the values
of
x for which the first four terms of this
expansion give a good approximation to the value of the function.Online
Calculate the value of x.
|
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-0.017536887899041176,
-0.022897208109498024,
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] |
169
The binomial expansion
8 a Find the first 4 terms
, in ascending powers of x, of the binomial expansion of (1 − 3x)5.
Give each term in its simplest form. (4 marks)
b If x
is small, so that x2 and higher powers can be ignored, show that
(1 + x)(1 −3x)5 ≈ 1 − 14x. (2 marks)
9 A microchip company mode
ls the probability of having no faulty chips on a single production
run as:
P(no fault) = (1 − p)n, p < 0.001
where p is the probability of a single chip being faulty, and n being the total number of chips
produced.
a State wh
y the model is restricted to small values of p. (1 mark)
b Given tha
t n = 200, find an approximate expression for P(no fault) in the form
a + bp + cp2. (2 marks)
c The company wants to achie
ve a 92% likelihood of having no faulty chips on a production
run of 200 chips. Use your answer to part b to suggest a maximum value of p for this to
be the case. (4 marks)E
E/P
1 The 16th row of Pascal’s triangle is shown below.
1 15 105 … …
a Find the next two v
alues in the row.
b Hence find the coefficient of x3 in the expansion of (1 + 2x)15.
2 Given tha
t ( 45 17 ) = 45 ! _____ 17 !a ! , write down the value of a. (1 mark)
3 20 people play a game at a school fete
.
The probability that exactly n people win a prize is modelled as ( 20 n ) p n (1 − p) 20 − n , where p is the
probability of any one person winning.
Calculate the probability of:a
5 people winning when p
= 1 _ 2
b nobody winning when
p = 0.7
c 13 people winning when p
= 0.6
Give your answers to 3 significant figures.
4 When (1 − 3 _ 2 x) p is expanded in ascending powers of x, the coefficient of x is −24.
a Find the value of
p. (2 marks)
b Find the coefficient of
x2 in the expansion. (3 marks)
c Find the coefficient of
x3 in the expansion. (1 mark)
5 Given tha
t:
(2 − x)13 ≡ A + Bx + Cx2 + …
find the values of the integers A, B and C. (4 marks)P
E
E/P
E/PMixed exercise 8
|
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] |
170
Chapter 8
6 a Expand (1 − 2
x)10 in ascending powers of x up to and including the term in x3,
simplifying each coefficient in the expansion. (4 marks)
b Use your expansion to f
ind an approximation of 0.9810, stating clearly the
substitution which you have used for x. (3 marks)
7 a Use the binomial series to expand (2
− 3x)10 in ascending powers of x up to and
including the term in x3, giving each coefficient as an integer. (4 marks)
b Use your series expansion, with a suita
ble value for x , to obtain an estimate for 1.9710,
giving your answer to 2 decimal places. (3 marks)
8 a Expand (3 + 2
x)4 in ascending powers of x, giving each coefficient as an integer. (4 marks)
b Hence, or otherwise, write do
wn the expansion of (3 − 2x)4 in ascending
powers of x. (2 marks)
c Hence by choosing a suitab
le value for x show that (3 + 2 √ _
2 ) 4 + (3 − 2 √ _
2 ) 4 is an
integer and state its value. (2 marks)
9 The coefficient of
x2 in the binomial expansion of (1 + x __ 2 ) n, where n is a positive
integer, is 7.
a Find the value of
n. (2 marks)
b Using the value of
n found in part a, find the coefficient of x4. (4 marks)
10 a Use the binomial theorem to e
xpand (3 + 10x)4 giving each coefficient as
an integer. (4 marks)
b Use your expansion, with an a
ppropriate value for x, to find the exact value of
10034. State the value of x which you have used. (3 marks)
11 a Expand (1 + 2
x)12 in ascending powers of x up to and including the term in x3,
simplifying each coefficient. (4 marks)
b By substituting a suitable v
alue for x, which must be stated, into your answer to
part a, calculate an approximate value of 1.0212. (3 marks)
c Use your calcula
tor, writing down all the digits in your display, to find a more exact
value of 1.0212. (1 mark)
d Calculate
, to 3 significant figures, the percentage error of the approximation found
in part b. (1 mark)
12 Expand (x − 1 __ x ) 5, simplifying the coefficients. (4 marks)
13 In the binomial expansion of
(2k + x)n, where k is a constant and n is a positive integer, the
coefficient of x2 is equal to the coefficient of x3.
a Prov
e that n = 6k + 2. (3 marks)
b Given a
lso that k = 2 _ 3 , expand (2k + x)n in ascending powers of x up to and including
the term in x3, giving each coefficient as an exact fraction in its simplest form. (4 marks)E
E
E/P
E/P
E
E
E/P
E/P
|
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] |
171
The binomial expansion
14 a Expand (2 +
x)6 as a binomial series in ascending powers of x, giving each coefficient
as an integer. (4 marks)
b By making suitab
le substitutions for x in your answer to part a, show that
(2 + √ _
3 )6 − (2 − √ _
3 )6 can be simplified to the form k √ __
3 , stating the value of the
integer k. (3 marks)
15 The coefficient of
x2 in the binomial expansion of (2 + kx)8, where k is a positive constant,
is 2800.
a Use algebr
a to calculate the value of k. (2 marks)
b Use your va
lue of k to find the coefficient of x3 in the expansion. (4 marks)
16 a Given tha
t
(2 + x)5 + (2 − x)5 ≡ A + Bx2 + Cx4,
find the value of the constants A, B and C. (4 marks)
b Using the substitution y
= x2 and your answers to part a, solve
(2 + x)5 + (2 − x)5 = 349. (3 marks)
17 In the binomial expansion of
(2 + px)5, where p is a constant, the coefficient of x3 is 135.
Calculate:a
the value of
p, (4 marks)
b the value of
the coefficient of x4 in the expansion. (2 marks)
18 Find the constant term in the expansion of
( x 2 ___ 2 − 2 __ x ) 9
.
19 a Find the first three terms
, in ascending powers of x of the binomial expansion of
(2 + px)7, where p is a constant. (2 marks)
The first 3 terms are 128, 2240
x and qx2, where q is a constant.
b Find the value of
p and the value of q. (4 marks)
20 a Write down the first thr
ee terms, in ascending powers of x, of the binomial expansion
of (1 − px)12, where p is a non-zero constant. (2 marks)
b Given tha
t, in the expansion of (1 − px)12, the coefficient of x is q and the coefficient
of x2 is 6q, find the value of p and the value of q. (4 marks)
21 a Find the first 3 terms, in ascending po
wers of x, of the binomial expansion
of (2 + x __ 2 ) 7
, giving each term in its simplest form. (4 marks)
b Explain how you w
ould use your expansion to give an estimate for the value of 2.057. (1 mark)
22 g(x
) = (4 + kx)5, where k is a constant.
Given that the coefficient of x3 in the binomial expansion of g(x) is 20,
find the value of k. (3 marks)E/P
E/P
E/P
E/P
P
E/P
E/P
E/P
E/P
|
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] |
172
Chapter 8
1 Pascal’s triangle is formed by adding adjacent pairs of numbers to find the numbers on the
next r
ow.
2 The (n
+ 1)th row of Pascal’s triangle gives the coefficients in the expansion of (a + b)n.
3 n! =
n × (n − 1) × (n − 2) × … × 3 × 2 × 1.
4 You can use factorial notation and y
our calculator to find entries in Pascal’s triangle quickly.
• The number of wa
ys of choosing r it
ems from a group of n items is
written as nCr or ( n r ) : nCr = ( n r ) = n ! ________ r ! (n − r) !
• The rth ent
ry in the nth row of Pascal’s triangle is given by n − 1Cr − 1 = ( n − 1 r − 1 ) .
5 The binomial expansion is:
(a + b) n = a n + ( n 1 ) a n − 1 b + ( n 2 ) a n − 2 b 2 + . . . + ( n r ) a n − r b r + . . . + b n (n ∈ ℕ)
where ( n r ) = n C r = n ! _________ r !(n − r) !
6 In the expansion o
f (a + b)n the general term is given by ( n r ) an − rbr.
7 The first few terms in the binomial e
xpansion can be used to find an approximate value for a
complicated expression.Summary of key points1 f(x) = (2 − px)(3 + x )5 where p is a constant.
There is no x2 term in the expansion of f( x).
Show that p = 4 _ 3
2 Fin
d the coefficient of x2 in the expansion of (1 + 2 x)8(2 − 5x)7.Challenge
|
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] |
173
Trigonometric ratios
After completing this unit you should be able to:
● Use the cosine rul
e to find a missing side or angle → pages 174–179
● Use the sine rule to find a missing side or angl e → pages 179–185
● Find the area of a t riangle using an appropriate
formula → pages 185–187
● Solve problems involving triangles → pages 187–192
● Sketch the graphs of the sine, cosine and tangent
functions → pages 192–194
● Sketch simple transformations of these graphs → pages 194–198Objectives
1 Use trigonometry to find the lengths of the marked sides
.
a
x
23°
7.3 cm b x
70°
8.5 cm
←GCSE Mathematics
2 Find the sizes of the angl es marked.
a
6.2 cm2.7 cmθ b 5 cm
22 cmθ
← GCSE Mathematics
3 f(x) = x2 + 3x. Sketch the graphs of
a y =
f(x) b y =
f(x + 2)
c y =
f(x) − 3 d y =
f( 1 _ 2 x)
← Sections 4.5, 4.6Prior knowledge check
Trigonometry in both two and three
dimensions is used by surveyors to work out distances and areas when planning building projects. You will also use trigonometry when working with vector quantities in mechanics.
→ Exercise 9B Q12 and Mixed exercise Q10, Q119
|
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] |
174
Chapter 9
9.1 The cosine rule
The cosine rule can be used to work out missing sides or angles in triangles.
■ This version of the c
osine rule is used to find a missing side if you know two sides and the
angle between them:
a2 = b2 + c2 − 2bc cos A
A
CB
ac
b
You can use the standard trigonometric ratios for right-angled triangles to prove the cosine rule: You c an exchange the letters depending
on which side you want to find, as long as each side
has the same letter as the opposite angle.Watch out
A BXba
h
xc – xC
h2 + x2 = b2
and h2 + (c − x)2 = a2
So x2 − (c − x)2 = b2 − a2
So 2cx − c2 = b2 − a2
a2 = b2 + c2 − 2 cx (1)
bu
t x = b
cos A (2)
So a2 = b2 + c2 − 2 bc cos AUse Pythagoras’ theorem in △CAX.
Use Pythagoras’ theorem in △CBX.
Subtract the two equations.
(c − x)2 = c2 − 2cx + x2.
So x2 − (c − x)2 = x2 − c2 + 2cx − x2.
Rearrange.Use the cosine ratio cos A =
x __ b in △CAX.
Combine (1) and (2). This is the cosine rule. For a right-angled triangle
Hypotenuse
AdjacentOpposite
θHint
sin θ = opposite __________ hypotenuse
cos
θ = adj
acent __________ hypotenuse
tan
θ = opp
osite ________ adjacent
If you are given all three sides and asked to find an angle, the cosine rule can be rearranged.
a2 + 2bc cos A = b2 + c2
2bc cos A = b2 + c2 − a2
Hence cos A = b2 + c2 − a2 __________ 2bc
You can ex
change the letters depending on which angle you want to find.
■ This version of the cosine rule is used to find an angle if you know all three sides:
cos A = b2 + c2 − a2 __________ 2bc Explore the cosine rule using
GeoGe
bra.Online
|
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] |
175
Trigonometric ratios
Example 1
Calculate the length of the side AB of the triangle ABC in which AC = 6.5 cm, BC = 8.7 cm and
∠ACB
= 100°.
A
B Cb = 6.5 cm
a = 8.7 cm100°c
c2 = a2 + b2 − 2 ab cos C
c2 = 8.72 + 6.52 − 2 × 8.7 × 6.5 × cos 100°
= 75.69 + 42.25 − ( −19.639…)
= 137. 57…
So c = 11.729…
So AB = 11.7 cm (3 s
.f.)Write out the formula you are using as the first line
of working, then substitute in the values given.
Don’t round any values until the end of your working. You can write your final answer to 3 significant figures.
Find the square root.Label the sides of the triangle with small letters a, b and c opposite the angles marked.
Example 2
Find the size of the smallest angle in a triangle whose sides have lengths 3 cm, 5 cm and 6 cm.
A BC
3 cm5 cm6 cm
cos C = a2 + b2 − c2 ___________ 2ab
cos
C = 52 + 62 − 32 ____________ 2 × 5 × 6
= 0.
8666…
C = 29.9° (3 s.f.)
The size of the smallest angle is 29.9°.Label the triangle ABC.
The smallest angle is opposite the smallest side so angle C is required.
Use the cosine rule cos C = a2 + b2 − c2 __________ 2ab
C = cos−1(0.8666…) Use your calculator to work this
ou
t efficiently.Online
|
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] |
176
Chapter 9
Example 3
Coastguard station B is 8 km, on a bearing of 060°, from coastguar d station A. A ship C is 4.8 km,
on a bearing of 018°, aw
ay from A. Calculate how far C is from B.
ABCa km
4.8 km
8 km18°
60°N
a2 = b2 + c2 − 2 bc cos A
a2 = 4.82 + 82 − 2 × 4.8 × 8 × cos 42°
= 29.966…
a = 5.47 (3 s.f.)
C is 5.47 km aw
ay from coastguard B .You now have b = 4.8 km, c = 8 km and A = 42°.
Use the cosine rule a2 = b2 + c2 − 2bc cos A.
If possible, work this out in one go using your
calculator.
Take the square root of 29.966… and round your final answer to 3 significant figures.If no diagram is given with a question you should draw one carefully. Double-check that the information given in the question matches your sketch.Problem-solving
In △ABC, ∠CAB = 60° − 18° = 42°.
Example 4
In △ABC, AB = x cm, BC = (x + 2) cm, AC = 5 cm and ∠ABC = 60°.
Find the value of x.
A
B Cx cm
(x + 2) cm5 cm
60°
b2 = a2 + c2 − 2 ac cos B
52 = (x + 2)2 + x2 − 2x(x + 2) cos 60°
So 25 = 2x2 + 4 x + 4 − x2 − 2x
So x2 + 2x − 21 = 0
x = −2 ± √ ___ 88 __________ 2
= 3.
69 (3 s.f.)Use the information given in the question to draw
a sketch.
Carefully expand and simplify the right-hand side. Note that cos 60° =
1 _ 2
Rearrange to the form ax2 + bx + c = 0.
Solve the quadratic equation using the quadratic
formula. ← Section 2.1
x represents a length so it cannot be negative.
|
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] |
177
Trigonometric ratios
Give answers to 3 significant figures, where appropriate.
1 In each of the follo
wing triangles calculate the length of the missing side.
a
20°
AB
C
8.4 cm6.5 cm b
60°AB
C2 cm
1 cm c
160°A
BC4.5 cm
5.5 cm
d
45°AB
C6 cm5 cm e
40°
AB
C10 cm10 cm f
108°AB
C5.6 cm6.5 cm
2 In the following triangles calculate the size of the angle marked x:
a
AB
C10 cm8 cmx
4 cm b
AB
C
24 cm25 cm7 cm
x c B
A
C4 cm3.5 cm2.5 cm
x
d
AB
C10 cm8 cm 7 cm
x e A
BC9 cm6 cm14 cm
x f
xAB
C6.2 cm6.2 cm3.8 cm
3 A plane flies from airport A on a bearing of 040° for 120 km
and then on a bearing of 130° f
or 150 km. Calculate the
distance of
the plane from the airport.
A120 km 150 km40°130°NNExercise 9A
|
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] |
178
Chapter 9
4 From a point
A a boat sails due north for 7 km to B. The boa
t leaves B and moves on a bearing
of 100° for 10 km until it reaches C
. Calculate the distance of C from A.
5 A helicopter flies on a bearing of 080° fr
om A to B, where AB = 50 km.
It then flies for 60 km to a point C.
Gi
ven that C is 80 km from A
, calculate the bearing of C from A.
6 The distance from the tee,
T, to the flag, F, on a particular hole on a golf course is 494 yards.
A golfer’s tee shot travels 220 yards and lands at the point S, where ∠STF = 22°.
Calculate how far the ball is from the flag.
7 Show that cos
A = 1 _ 8 A
B C4 cm 5 cm
6 cm
8 Show that cos
P = − 1 _ 4
RP Q
4 cm3 cm2 cm
9 In △ABC, AB = 5 cm, BC = 6 cm and AC = 10 cm.
Calcula
te the size of the smallest angle.
10 In △ABC
, AB = 9.3 cm, BC
= 6.2 cm and AC =
12.7 cm.
Calcula
te the size of the largest angle.
11 The lengths of the sides of a triangle ar
e in the ratio 2 : 3 : 4.
Calcula
te the size of the largest angle.
12 In △ABC
, AB = (x − 3) cm, BC
= (x + 3) cm, AC =
8 cm and ∠BA
C = 60°.
Use the cosine rule to find the value of x.
13 In △ABC
, AB = x cm, BC
= (x − 4) cm, AC =
10 cm and ∠BA
C = 60°.
Calculate the value of x.
14 In △ABC
, AB = (5 − x) cm, BC
= (4 + x) cm, ∠ABC
= 120° and AC = y cm.
a Show that
y2 = x2 − x + 61.
b Use the method of completing the square to find the minim
um value of y2, and give the
value of x for which this occurs.P
P
P
P
P
|
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] |
179
Trigonometric ratios
15 In △ABC
, AB = x cm, BC
= 5 cm, AC =
(10 − x) cm.
a Show that cos
∠ABC = 4x
− 15 _______ 2x
b Given tha
t cos ∠ABC = − 1 _ 7 , work out the value of x.
16 A farmer has a field in the sha
pe of a quadrilateral as shown.
D
ABC
75 m60 m
120 m135 m
The angle between fences AB and AD is 74°. Find the angle between fences BC and CD.
17 The diagram sho
ws three cargo ships, A, B and C, which are in
70 km
50 km
20°N
B
AC
the same horizontal plane. Ship B is 50 km due north of ship A
and ship C is 70 km from ship A
. The bearing of C from A is 020°.
a Calculate the distance betw
een ships B and C, in kilometres
to 3 s.f. (3 marks)
b Calculate the bearing of
ship C from ship B. (4 marks)P
P
You will have to use the cosine rule twice.
Copy the diagram and write any angles or lengths you work out on your copy.Problem-solving
E/P
9.2 The sine rule
The sine rule can be used to work out missing sides or angles in triangles.
■ This version of the sine rule is used to find the length A
CB
ac
b
of a missing side:
a _____ sin A = b _____ sin B = c _____ sin C
You can use the standard t
rigonometric ratios for right-angled triangles to prove the sine rule:
ABXa bhC
c
sin B = h __ a ⇒ h = a sin BIn a general triangle ABC, draw the perpendicular
from C to AB. It meets AB at X.
The length of CX is h.
Use the sine ratio in triangle CBX.
Explore the sine rule using
GeoGe
bra.Online
|
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] |
180
Chapter 9
and sin A = h __ b ⇒ h = b sin A
So a sin
B = b
sin A
So a _____ sin A = b _____ sin B
In a si
milar way, by drawing the perpendicular
from B to the side AC , you can show that:
a _____ sin A = c _____ sin C
So a _____ sin A = b _____ sin B = c _____ sin C Use the sine ratio in triangle CAX.
Divide throughout by sin A sin B .
This is the sine rule and is true for all triangles.
■ This version of the sine rule is used to find a missing angle:
sin A _____ a = sin B _____ b = sin C _____ c
Example 5
In △ABC, AB = 8 cm, ∠BA C = 30° and ∠BCA = 40°. Find BC.
ACB
x cm 8 cm
30° /four.ss010°
a _____ sin A = b _____ sin B
x _______ sin 30° = 8 _______ sin 40°
So x = 8 sin 3
0° _________ sin 40° = 6.2228…
= 6.
22 cm (3 s
.f.)Always draw a diagram and carefully add the data.
Here c = 8 (cm), C = 40°, A = 30°, a = x (cm).
In a triangle, the larger a side is, the larger the opposite angle is. Here, as C > A, then c > a, so
you know that 8 > x.
Multiply throughout by sin 30°.
Give your answer to 3 significant figures.Use this version of the sine rule to find a missing side. Write the formula you are going to use as the first line of working.
Example 6
In △ABC, AB = 3.8 cm, BC = 5.2 cm and ∠ABC = 35°. Find ∠ABC.
3.8 cm 5.2 cm
A CB
35°Here a = 5.2 cm, c = 3.8 cm and A = 35°.
You first need to find angle C.
|
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] |
181
Trigonometric ratios
sin C _____ c = sin
A _____ a
sin
C _____ 3.8 = sin
35° _______ 5.2
So si
n C = 3.8
sin 35° ___________ 5.2
C = 24
.781…
So B = 12
0° (3 s.f.)Use sin C _____ c = sin A _____ a
Write the f
ormula you are going to use as the first
line of working.
Use your calculator to find the value of C in a
single step. Don’t round your answer at this point.
B = 180° − (24.781…° + 35°) = 120.21… which rounds to 120° to 3 s.f.
Give answers to 3 significant figures, where appropriate.
1 In each of parts a
to d, the given values refer to the general triangle.
AB
ca
bCa Given that a = 8 cm, A = 30°, B = 72°, find b.
b Given tha
t a = 24 cm, A
= 110°, C = 22°, find c.
c Given tha
t b = 14.7 cm, A
= 30°, C = 95°, find a.
d Given tha
t c = 9.8 cm, B
= 68.4°, C = 83.7°, find a.
2 In each of the follo
wing triangles calculate the values of x and y.
x
60°57°xab
c7.5 cm25 cm
8 cmy cm
y cm
y cmx cm
39° 30°
145°112°
56.4°36.8°
53.2°72°xd
ef8 cm
5.9 cm6 cmy cm
y cm
x cm
x cm
y cmy cm50°Exercise 9B
In parts c and d ,
st
art by finding the
size of the third angle.Hint
|
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] |
182
Chapter 9
3 In each of the follo
wing sets of data for a triangle ABC,
AB
ca
bC
find the value of x.
a AB =
6 cm, BC
= 9 cm, ∠BA
C = 117°, ∠ACB = x
b AC =
11 cm, BC
= 10 cm, ∠ABC
= 40°, ∠CAB = x
c AB =
6 cm, BC
= 8 cm, ∠BA
C = 60°, ∠ACB = x
d AB =
8.7 cm, AC =
10.8 cm, ∠ABC
= 28°, ∠BAC = x
4 In each of the diagr
ams shown below, work out the size of angle x.
x50°
xx
AAAC C CBB B abc
5.8 cm 4.5 cm3 2 cm2 cm
6.2 cm7.2 cm
67.5°80°
x
x55°
60°x A
AA
CC
CB B B de f
10 cm
7.9 cm
10.4 cm9.7 cm12.4 cm
8 cm
70°
5 In △PQR, QR = √ __
3 cm, ∠PQR = 45° and ∠QPR = 60°. Find a PR and b PQ.
6 In △PQR
, PQ = 15 cm, QR
= 12 cm and ∠P
RQ = 75°. Find the two remaining angles.
7 In each of the follo
wing diagrams work out the values of x and y.
y100°x
x
AA C C
DB
ACB B ab c
3.9 cm8.5 cm
12.2 cm7 cm
20°x cm 10.8 cm
y cmy cm5.5 cm
75°45°
78°80°yx
ACC
DA ADDCBB B de f
6.4 cm 7 cm
xxy cmy cm
6 cm7.5 cm 6.2 cm
8.5 cm
5 cm10°
60° 24°
8 Town B is 6 km, on a bearing of 020°, from to wn A.
Town C is located on a bearing of 055° from town A
and on a bearing of 120° from town B. Work out the distance of town C from: a
town A
b town BP
Draw a sketch to show the information.Problem-solving
|
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] |
183
Trigonometric ratios
9 In the diagram
AD = DB = 5 cm, ∠ABC
= 43°
43°5 cm
5 cmA
CDB72°
and ∠ACB = 72°.
Calculate:
a AB
b CD
10 A zookeeper is building an enclosur
e for some llamas. 66°
76 m
80 m98°A B
CD
The enclosure is in the shape of a quadrilateral as shown.
If the length of the diagonal BD is 136 m
a find the angle between the fences AB
and BC
b find the length of fence AB
11 In △ABC
, AB = x cm, BC
= (4 − x) cm,
∠BA
C = y and ∠BCA = 30°.
Given that sin y = 1 ___ √ __
2 , show that
x
= 4( √ __
2 − 1) (5 marks)
12 A surveyor w
ants to determine the height of a
15 m40° 52°
building. She measures the angle of elevation
of the top of the building at two points 15 m
apart on the gr
ound.
a Use this informa
tion to determine the height
of the building. (4 marks)
b State one assumption made by the surv
eyor
in using this mathematical model. (1 mark)
■ The sine rule sometimes produc es two possible
solutions for a missing angle:
• sin θ = sin (180° − θ )E/P
You can use the value of sin y directly in your
calculation. You don’t need to work out the value of y .Problem-solving
E/P
For given side lengths b and c and given angle B, you can draw the triangle in two different ways.
b bcA
BC1 C2You can draw b such that the angle at C is obtuse (C
1).Alternatively you can draw b such that the angle at C is acute (C
2).
Since AC1C2 is an isosceles triangle, it follows that the
angles AC1B and AC2B add together to make 180°. You can confirm this relationship by
co
nsidering the graph of y = sin x .
sin θ
180 – θ180° θ1
xy
O
→ Section 9.5 and Chapter 10Links
|
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] |
184
Chapter 9
Example 7
In △ABC, AB = 4 cm, AC = 3 cm and ∠ABC = 44°. Work out the two possible values of ∠ACB .
Give answers to 3 significant figures, where appropriate.
1 In △ABC
, BC = 6 cm, AC =
4.5 cm and ∠ABC
= 45°.
a Calculate the tw
o possible values of ∠BAC .
b Draw a dia
gram to illustrate your answers.
2 In each of the diagr
ams shown below, calculate the possible values of x and the corresponding
values of y.
x
50°
40°A
B
CAB
C AB
Cy
xx12 cm8 cm
y cm
y cm25.6°42 cm
21 cm5 cm 4 cmab c
3 In each of the follo wing cases △ABC has ∠ABC = 30° and AB = 10 cm.
a Calculate the least possib
le length that AC could be.
b Given tha
t AC = 12 cm, calculate
∠ACB .
c Given instead tha
t AC = 7 cm, calculate the tw
o possible values of ∠ACB .Exercise 9CA
BC3 cm
C1 C23 cm/four.ss01 cm
/four.ss01/four.ss01°
sin C _____ c = sin
B _____ b
sin
C _____ 4 = sin
44° _______ 3
sin
C = 4 sin 4
4° _________ 3
So C = 67.
851… = 67.9° (3 s.f.)
or C = 180
− 67.851… = 112
.14…
= 112° (3 s.f.)Use sin C _____ c = sin B _____ b , where b = 3, c = 4, B = 44°.
As sin (180 − θ ) = sin θ .Think about which lengths and angles are fixed,
and which ones can vary. The length AC is fixed.
If you drew a circle with radius 3 cm and centre A it would intersect the horizontal side of the triangle at two points, C
1 and C2.Problem-solving
|
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] |
185
Trigonometric ratios
4 Triangle ABC
is such that AB = 4 cm, BC
= 6 cm and ∠AC
B = 36°. Show that one of the possible
values of ∠ABC is 25.8° (to 3 s.f.). Using this value, calculate the length of AC .
5 Two triangles
ABC are such that AB = 4.5 cm, BC
= 6.8 cm and ∠AC
B = 30°. Work out the value
of the largest angle in each of the triangles.
6 a A crane arm
AB of length 80 m is anchored at
point
B at an angle of 40° to the horizontal.
A wrecking ball is suspended on a cable of
length 60 m from A
. Find the angle x through
which the wrecking ball rotates as it passes the two points level with the base of the crane arm at B.
(6 marks)
b Write down one mode
lling assumption you
have made. (1 mark)P
P
E/P
x°
80 m
60 m
40°A
B
9.3 Areas of triangles
You need to be able to use the formula for finding the area of any triangle when you know two sides and the angle between them.
■ Area = 1 _ 2 ab sin C
A
CB
ac
b
A proof of the formula: As with the cosine rule, the letters are
in
terchangeable. For example, if you know angle B
and sides a and c , the formula becomes Area = 1 _ 2 ac sin B.Hint
B CXb
hA
Area of △ABC = 1 __ 2 ah (1)
B
ut h = b
sin C (2)
S
o Ar
ea = 1 __ 2 ab si n CThe perpendicular from A to BC is drawn and it
meets BC at X. The length of AX = h.
Area of triangle = 1 _ 2 base × height.
Use the sine ratio sin
C = h __ b in △AXC .
Substitute (2) into (1).
|
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] |
186
Chapter 9
Example 8
Work out the area of the triangle shown below.
B
C A75°
6.9 cm4.2 cm
Area = 1 __ 2 bc sin A
A
rea of △ ABC = 1 __ 2 × 6.9 × 4.2 × sin 75° cm2
= 14.0 cm2 (3 s.f.)Here b = 6.9 cm, c = 4.2 cm and angle A = 75°,
so use:
Area = 1 _ 2 bc sin A .
1 Calculate the area of each triangle.
80°B
BB A
AC
CC
A8.6 cm
7.8 cm2.5 cm6.4 cm 6.4 cm
3.5 cm
45°100°ab cExercise 9DExample 9
In △ABC, AB = 5 cm, BC = 6 cm and ∠ABC = x. Given that the area of △ABC is 12 cm2 and that
AC is the longest side, find the value of x.
xB
C A5 cm 6 cm
Area = 1 __ 2 ac sin B
Area △ABC =
1 __ 2 × 6 × 5 × sin x cm2
So 12 = 1 __ 2 × 66 × 5 × sin x cm2
So sin x = 0.8
x = 126.
86…
= 127
° (3 s.f.)Here a = 6 cm, c = 5 cm and angle B = x, so use:
Area = 1 _ 2 ac sin B .
Area of △ABC is 12 cm2.
sin x = 12 __ 15
There are two values of x for which sin x = 0.8,
53.13…° and 126.86…°, but here you know B is the largest angle because AC is the largest side.Problem-solving Explore the area of a triangle
us
ing GeoGebra.Online
|
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] |
187
Trigonometric ratios
2 Work out the possib
le sizes of x in the following triangles.
x
x30 cm
6.5 cm8.5 cm
8 cm6 cm
x40 cm
Area =
400 cm2Area =
12.4 cm2Area =
12 3 cm2ab c B
ACBBC
A
AC
3 A fenced triangular plot of ground has ar ea 1200 m2. The fences along the two smaller sides are
60 m and 80 m respectiv
ely and the angle between them is θ. Show that θ = 150°, and work out
the total length of fencing.
4 In triangle ABC,
BC = (x + 2) cm, A
C (x + 2) cm150°x cm
B
AC = x cm and ∠
BCA = 150°.
Given that the area of the triangle
is 5 cm2, work out the value of x,
giving your answer to 3 significant figures.
5 In △PQR
, PQ = (x + 2) cm, PR
= (5 − x) cm and ∠
QPR = 30°.
The area of the triangle is A cm2.
a Show that
A = 1 _ 4 (10 + 3 x − x2). (3 marks)
b Use the method of completing the square
, or otherwise, to find the maximum
value of A, and give the corresponding value of x. (4 marks)
6 In △ABC
, AB = x cm, AC =
(5 + x) cm
and ∠
BAC = 150°. Given that the area of the
triangle is 3 3 _ 4 cm2
a Show that x satisfies the equation x2 + 5x − 15 = 0. (3 marks)
b Calculate the v
alue of x, giving your answer to 3 significant figures. (3 marks)P
E/P
E/P
x represents a length so it must be positive.Problem-solving
9.4 Solving triangle probl ems
You can solve problems involving triangles by using the sine and cosine rules along with Pythagoras’
theorem and standard right-angled triangle trigonometry.
If some of the triangles are right-angled, try to use basic trigonometry and Pythagoras’ theorem first
to work out other information.
If you encounter a triangle which is not right-angled,
you will need to decide whether to use the sine rule or the cosine rule. Generally, use the sine rule when you are considering two angles and two sides and the cosine rule when you are considering three sides and one angle. The s ine rule is often
easier to use than the cosine rule. If you
know one side and an opposite angle in a triangle, try to use the sine rule to find other missing sides and angles.Watch out
|
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] |
188
Chapter 9
For questions involving area, check first whether you can use Area = 1 _ 2 × b ase × height, before using
the formula involving sine.
• to find an unknown angl
e given two sides and one opposite angle, use the sine rule
• to find an unknown side giv
en two angles and one opposite side, use the sine rule
• to find an unknown angl
e given all three sides, use the cosine rule
• to find an unknown side giv
en two sides and the angle between them, use the cosine rule
• to find the area giv
en two sides and the angle between them, use Area = 1 _ 2 ab sin C
Example 10
The diagram below shows the locations of four mobile B
A
DC
140°80 m75 m
55°
phone masts in a field. BC = 75 m,
CD
= 80 m, angle BCD
= 55° and angle ADC = 140°.
In order that the masts do not interfere with each
other, they must be at least 70 m apart.
Giv
en that A is the minimum distance from D, find:
a the distance A is fr
om B
b the angle BAD
c the area enclosed by the f
our masts
B
A
DC
1/four.ss010° 80 m70 m75 m
55°
a BD2 = BC2 + CD2 − 2(BC)(CD)cos (∠BCD)
BD2 = 752 + 802 − 2(75)(80)cos 55°
BD2 = 5142.08…
BD = √ ___________ 5142.08… = 71 .708…
sin(∠BD
C) __________ BC = sin(∠B
CD) __________ BD
sin(∠BD
C) = sin(55°) × 75 _____________ 71.708 = 0.85675…
∠BDC = 58.954…
∠BDA = 140 − 58.954… = 81.045…Find BD first using the cosine rule.
Store this value in your calculator, or write down
all the digits from your calculator display.Split the diagram into two triangles. Use the information in triangle BCD to work out the
length BD . You are using three sides and
one angle so use the cosine rule .Problem-solving
You know a side and its opposite angle (BD and ∠BCD), so use the sine rule to calculate angle BDC.
Find BDA and store this value, or write down all
the digits from your calculator display.
|
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] |
189
Trigonometric ratios
AB2 = AD2 + BD2 − 2(AD)(BD)cos(∠BDA)
AB2 = 702 + 71.708…2
− 2(70)(71.708…)cos (81.045…)
AB2 = 8479.55…
AB = √ ___________ 8479.55… = 92 .084…
= 92.1 m (3 s.
f.)
b sin(∠B
AD) __________ BD = sin(∠B
DA) __________ AB
sin(∠B
AD) = sin(81.045…) × 71.708 _______________________ 92.084
= 0.
769…
∠BAD = 50.28… = 50.3° (3 s.f.)
c Are
a ABCD = area BCD + area BDA
Area ABC
D = 1 __ 2 (BC)(CD) sin (∠BC D)
+ 1 __ 2 (AB)(AD) sin (∠BA D)
Area ABC
D = 1 __ 2 (75)(80)sin (55 °)
+ 1 __ 2 (92.084…)(70)sin (50 .28…°)
Area ABCD = 2457.4… + 2479.0…
Area ABCD = 4936.4… = 4940 m2 (3 s.f.)AB is a length, so you are only interested in the
positive solution.
Use the sine rule to calculate angle BAD .
Alternatively you could have used the cosine rule with sides AB, BD and AD.
Use the area formula twice.
Try to use the most efficient method, and give answers to 3 significant figures.
1 In each triangle below find the v
alues of x, y and z.
30°x
x
84°y
xa
4.2 cm
5.7 cm
ABB
AC
A CB
C z cm56°48°b
14.6 cm20 cmy cm
y cmz cmc
z cm
120°
yxd
12.8 cm
6 cmz cm
130°A
AACCC BB
B
yxy
x
zze3 cm
5 cm6 cm8 cm
8 cm12 cmf
y
x
z cm40°10.5 cm9.5 cmg
z45°
yz x
4.8 cm9.6 cm
x cmh
12.3 cm
15.6 cmy cmi
20.4 cm
C
C
CAB
D
A
AB BExercise 9EYou can now use the cosine rule in triangle ABD to find AB.
Explore the solution step-by-step
us
ing GeoGebra.Online
|
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] |
190
Chapter 930°x
x
84°y
xa
4.2 cm
5.7 cm
ABB
AC
A CB
C z cm56°48°b
14.6 cm20 cmy cm
y cmz cmc
z cm
120°
yxd
12.8 cm
6 cmz cm
130°A
AACCC BB
B
yxy
x
zze3 cm
5 cm6 cm8 cm
8 cm12 cmf
y
x
z cm40°10.5 cm9.5 cmg
z45°
yz x
4.8 cm9.6 cm
x cmh
12.3 cm
15.6 cmy cmi
20.4 cm
C
C
CAB
D
A
AB B
2 In △ABC, ca lculate the size of the remaining angles, the lengths of the third side and the area
of the triangle given that
a △BAC
= 40°, AB = 8.5 cm and BC
= 10.2 cm
b △ACB
= 110°, AC = 4.9 cm and BC
= 6.8 cm
3 A hiker wa
lks due north from A and after 8 km reaches B
. She then walks a further 8 km on a
bearing of 120° to C
. Work out a the distance from A to C and b the bearing of C from A.
4 A helicopter flies on a bearing of 200° fr
om A to B, where AB = 70 km. It then flies on a
bearing of 150° from
B to C, where C is due south of A. Work out the distance of C from A.
5 Two r
adar stations A and B are 16 km apart and A
is due north of B. A ship is known to be on
a bearing of 150° from A and 10 km from B
. Show that this information gives two positions for
the ship, and calculate the distance between these two positions.
6 Find x in each of
the following diagrams:
50° xBB B
A CD
A CCD
ADab c
8 cm
5 cm7 cm
10 cm5 cm50° 60°
x cmx cm
6 cm2.4 cm8.6 cm
3.8 cm
7 In △ABC, AB = 4 cm, BC = (x + 2) cm and AC = 7 cm.
a Explain how you kno
w that 1 < x < 9. B
AC4 cm
7 cm(x + 2) cmb Work out the value of x and the area of the triangle
for the cases w
hen
i ∠ABC
= 60° and
ii ∠ABC
= 45°, giving your answers to 3 significant figures.
8 In the triangle, cos ∠
ABC = 5 _ 8
B AC
2 cm6 cm
(x + 1) cma Calculate the value of x.
b Find the area of triangle
ABC.P
P
P
P
P
|
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] |
191
Trigonometric ratios
9 In △ABC
, AB = √ __
2 cm, BC = √ __
3 cm and ∠BAC = 60°. Show that ∠ACB = 45° and find AC .
10 In △ABC
, AB = (2 − x) cm, BC
= (x + 1) cm
and ∠
ABC = 120°.
a Show that
AC 2 = x2 − x + 7.
b Find the value of
x for which AC has a
minimum value.
11 Triangle ABC
is such that BC = 5 √ __
2 cm, ∠ABC = 30° and ∠BAC = θ, where sin θ = √ __
5 ___ 8
Work out the length of
AC, giving your answer in the form a √ __
b , where a and b are integers.
12 The perimeter of △
ABC = 15 cm. Given tha
t AB = 7 cm and ∠BA
C = 60°, find the lengths of
AC and BC and the area of the triangle.
13 In the triangle ABC,
AB = 14 cm, BC
= 12 cm and CA
= 15 cm.
a Find the size of angle
C, giving your answer to 3 s.f. (3 marks)
b Find the area of triangle
ABC, giving your answer in cm2 to 3 s.f. (3 marks)
14 A flower bed is in the sha
pe of a B
A C4.2 m7.5 m
2.1 m 3.5 m5.9 m
D
quadrilateral as shown in the diagram.
a Find the sizes of angles
DAB and BCD.
(4 marks)
b Find the total area of
the flower bed.
(3 marks)
c Find the length of the diagona
l AC .
(4 marks)
15 ABCD is a square
. Angle CED is obtuse.
40°AB
D CE10 cm
8 cmFind the area of the shaded triangle. (7 marks)P
P
Complete the square for the expression
x2 − x + 7 to find the minimum value of
AC2 and the value of x where it occurs.Problem-solving
P
E
E/P
E/P
|
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