1. CHAPTER 6 COORDINATE GEOMETRY FORM 4
20
PAPER 1
1. A point T divides the line segment joining the points A(1, -2) and B(-5, 4) internally in the
ratio 2 : 1. Find the coordinates of point T.
[2 marks]
2. Diagram below shows a straight line PQ with the equation
3
x
+
5
y
= 1. The point Q lies
on the x-axis and the point P lies on the y-axis.
Find the equation of the straight line perpendicular to PQ and passing through the point Q.
[3 marks]
3. The line 8x + 4hy - 6 = 0 is perpendicular to the line 3x + y = 16. Find the value of h.
[3 marks]
4. Diagram below shows the straight line AB which is perpendicular to the straight line CB at
the point B.
The equation of the straight line CB is y = 3x  4. Find the coordinates of B.
[3 marks]
5. The straight line
14
x
+
m
y
= 1 has a y-intercept of 3 and is parallel to the straight line
y + nx = 0. Determine the value of m and of n.
x
P
Q
y
0
A(0,6) B
x
y
C
0

2. CHAPTER 6 COORDINATE GEOMETRY FORM 4
21
[3 marks]
6. Diagram below shows a straight line passing through A(2, 0) and B (0, 6).
a) Write down the equation of the straight line AB in the form
a
x
+
b
y
= 1.
[1 mark]
b) A point P(x, y) moves such that PA = PB. Find the equation of the locus of P.
[2 marks]
x
B(0, 6)
A(2, 0)
y
0

3. CHAPTER 6 COORDINATE GEOMETRY FORM 4
22
PAPER 2
1. Solutions to this question by scale drawing will not be accepted.
Diagram shows a straight line CD which meets a straight line AB at the point D. The point C
lies on the y-axis.
0
a) Write down the equation of AB in the form of intercepts. [1 mark ]
b) Given that 2AD = DB, find the coordinates of D. [2 marks]
c) Given that CD is perpendicular to AB , find the y-intercept of CD. [3 marks]
2. Solutions to this question by scale drawing will not be accepted.
In the diagram the straight line BC has an equation of 3y + x + 6 = 0 and is perpendicular to
straight line AB at point B.
(a) Find
i) the equation of the straight line AB
ii) the coordinates of B. [5 marks]
(b) The straight line AB is extended to a point D such that AB : BD = 2 : 3. Find the
coordinates of D. [2 marks]
(c) A point P moves such that its distance from point A is always 5 units.
Find the equation of the locus of P. [3 marks]
0
x
y
DA(0 , -3)
C
B (12, 0)
A(-6, 5)
B
C
3y + x + 6 = 0
x
y
0

4. CHAPTER 6 COORDINATE GEOMETRY FORM 4
23
3. Solutions to this question by scale drawing will not be accepted.
Diagram shows the triangle AOB where O is the origin. Point C lies on the straight line AB.
(a) Calculate the area, in unit2
, of triangle AOB. [2 marks]
(b) Given that AC : CB = 3 : 2, find the coordinates of C. [2 marks]
(c) A point P moves such that its distance from point A is always twice its distance from point
B.
(i) Find the equation of the locus of P.
(ii) Hence, determine whether or not this locus intercepts the y-axis. [6 marks]
4. In the diagram, the straight line PQ has an equation of y + 3x + 9 = 0. PQ intersects the
x-axis at point P and the y-axis at point Q.
Point R lies on PQ such that PR : RQ = 1 : 2. Find
(a) the coordinates of R, [3 marks]
(b) the equation of the straight line that passes through R and perpendicular to PQ.
[3 marks]
y + 3x + 9 = 0
y
x
A(-2, 5)
B(5, -1)
0
C
y
x
P
Q
0
R

5. CHAPTER 6 COORDINATE GEOMETRY FORM 4
24
5. Solutions to this question by scale drawing will not be accepted.
Diagram shows the triangle OPQ. Point S lies on the line PQ.
a) A point W moves such that its distance from point S is always 2
2
1
units.
Find the equation of the locus of W. [3 marks]
b) It is given that point P and point Q lie on the locus of W.
Calculate
i) the value of k,
ii) the coordinates of Q.
[5 marks]
c) Hence, find the area , in unit2
, of triangle OPQ.
[2 marks]
0
x
y P(3 , k)
S (5, 1)
Q

6. CHAPTER 6 COORDINATE GEOMETRY FORM 4
25
ANSWERS ( PAPER 1 )
1.
T (
3
)2)(5()1)(1( 
,
3
)2)(4()1)(2( 
)
2
= T( -3 , 2 ) 1
2.
Gradient of PQ , m1 = -
3
5
and the coordinates of Q (3 , 0)
1
Let the gradient of straight line perpendicular to PQ and passing through Q
= m2 . Then m1  m2 = -1.
m2 =
5
3
 The equation of straight line is
3
0


x
y
=
5
3
5y = 3(x – 3)
1
5y = 3x – 9 1
3. Given 8x + 4hy – 6 = 0
4hy = -8x + 6
y = -
h4
8
x +
h4
6
y = -
h
2
x +
h2
3
Gradient , m1 = -
h
2
3x + y = 16
y = -3x + 16
Gradient , m2 = -3
1
Since the straight lines are perpendicular to each other , then m1  m2 = -1.
 (-
h
2
)(-3) = -1
1
6 = -h
h = -6 1
4. Gradient of CB , m1 = 3
Since AB is perpendicular to CB, then m1 m2 = 1
Gradient of AB, m2 = 
3
1 1
 The equation of AB is y = -
3
1
x + 6
B is the point of intersection.
y = 3x  4 ……………(1)
y = 
3
1
x + 6 ……………(2)
3x  4 = 
3
1
x + 6
1

7. CHAPTER 6 COORDINATE GEOMETRY FORM 4
26
3
10
x = 10
x = 3
y = 3(3)  4
= 5
The coordinates of B are (3, 5). 1
5.
14
x
+
m
y
= 1
 y-intercept = m = 3
1
From
14
x
+
3
y
= 1, the gradient m1 = -
14
3
From y = -nx , the gradient m2 = -n .
Since the two straight lines are parallel , then m1 = m2
-
14
3
= -n 1
 n =
14
3
1
6. a) From the graph given, x- intercept = 2 and y-intercept = 6.
The equation of AB is
2
x
+
6
y
= 1 . 1
b) Let the coordinates of P = (x , y) and since PA = PB
22
)0()2(  yx = 22
)6()0(  yx
(x – 2)2
+ y2
= x2
+ (y – 6)2
x2
– 4x + 4 + y2
= x2
+ y2
– 12y + 36
1
12y – 4x -32 = 0
3y – x - 8 = 0 1

8. CHAPTER 6 COORDINATE GEOMETRY FORM 4
27
ANSWERS ( PAPER 2 )
1
a)
12
x
-
3
y
= 1
1
b) Given 2AD = DB , so
DB
AD
=
2
1
 D = (
3
)1(12)2(0 
,
3
)1(0)2(3 
) 1
= ( 4 , -2 ) 1
c) Gradient of AB, mAB = -(
12
3
)
=
4
1
1
Since AB is perpendicular to CD, then mAB mCD = 1.
 Gradient of CD, mCD = - 4
Let, coordinates of C = (0 , h) ,
mCD =
40
)2(

h
- 4 =
4
2

h
16 = h + 2
h = 14
1
 y-intercept of CD = 14 1
2 a) i) Given equation of BC, 3y + x + 6 = 0
y = -
3
1
x – 2
Gradient of BC = -
3
1
1
Since AB is perpendicular to BC , then mAB mBC = 1.
Gradient of AB, mAB = 3
The equation of AB ,
)6(
5


x
y
= 3
y – 5 = 3x + 18
1
y = 3x + 23 1
ii) B is the point of intersection.
Equation of AB , y = 3x + 23 …………. (1)
Equation of BC , 3y + x + 6 = 0 ………….(2)
Substitute (1) into (2), 3(3x + 23) + x + 6 = 0
1

9. CHAPTER 6 COORDINATE GEOMETRY FORM 4
28
9x + 69 + x + 6 = 0
x = -
2
15
Substitute value of x into (1), y = 3(-
2
15
) + 23
y =
2
1
 The coordinates of B are ( -
2
15
,
2
1
)
1
b) Let D (h, k)
B( -
2
15
,
2
1
) = (
5
)18(2 h
,
5
152 k
) 1
-
2
15
=
5
)18(2 h
,
-75 = 4h – 36
h =
4
39
2
1
=
5
152 k
5 = 4k + 30
k =
4
25

1 The coordinates of D are (
4
39
,
4
25
 )
c) Given PA = 5
22
)5())6((  yx = 5 1
( x + 6)2
+ ( y – 5)2
= 25 1
x2
+ 12x + 36 + y2
-10y + 25 = 25
x2
+ y2
+ 12x -10y + 36 = 0 1
3 .)
a) Area =
2
1
0510
0250


=
2
1
)2()25( 
1
=
2
23
unit2 1
b) C = (
5
)2(2)5(3 
,
5
)5(2)1(3  1
= (
5
11
,
5
7
)
1
c) i) Since PA = 2PB
22
)5()2(  yx = 2 22
)1()5(  yx 1
x2
+ 4x + 4 + y2
10y + 25 = 4 (x2
 10x + 25 + y2
+2y + 1) 1

10. CHAPTER 6 COORDINATE GEOMETRY FORM 4
29
x2
+ y2
+ 4x 10y + 29 = 4x2
+ 4y2
40x + 8y + 104
3x2
+ 3y2
44x + 18y + 75 = 0 1
(ii) When it intersects the y-axis, x = 0.
 3y2
+1 8y + 75 = 0 1
Use b2
 4ac
= (18)2
 4(3)(75) 1
= 576
b2
 4ac < 0
It does not cut the y-axis since there is no real root. 1
4. a) y + 3x + 9 = 0
When y = 0, 0 + 3x + 9 = 0
x = –3
 P(–3, 0)
When x = 0, y + 0 + 9 = 0
y = –9
 Q(0, –9)
1
R(x, y) = (
3
)3(2)0(1 
,
3
)0(2)9(1 
)
1
= (-2 , -3 ) 1
b) y + 3x + 9 = 0
y = -3x - 9
 Gradient of PQ , m1 = –3
1
Since PQ is perpendicular to the straight line, then m1 m2 = 1
Thus,
3
1
2 m
The equation of straight line that passes through R(-2, -3) and
perpendicular to PQ is
2
3


x
y
=
3
1
1
3y = x - 7 1
5. a) Equation of the locus of W,
22
)1()5(  yx =
2
5
1
(x – 5)2
+ ( y – 1)2
= (
2
5
)2 1
x2
-10x +25 + y2
– 2y + 1 =
4
25
4 x2
+ 4y2
– 40x - 8y + 79 = 0 1
b) i) P(3 , k) lies on the locus of W,
substitute x =3 and y = k into the equation of the locus of W.
4(3)2
+ 4(k)2
– 40(3) – 8(k) + 79 = 0 1

11. CHAPTER 6 COORDINATE GEOMETRY FORM 4
30
4k2
- 8k -5 = 0
(2k + 1)(2k – 5) = 0
k = -
2
1
, k =
2
5
Since k > 0,  k =
2
5
1
1
ii) Since S is the centre of the locus of W, then S is the
mid-point of PQ.
S(5 , 1) = (
2
3x
,
2
2
5
y
) 1
5 =
2
3x
, 1 =
2
2
5
y
x = 7 , y = -
2
1
Hence, the coordinates of Q are ( 7 , -
2
1
). 1
c) Area of triangle OPQ =
2
1
0
2
5
2
1
0
0370

=
2
1
[ (7)(
2
5
) – (-
2
3
) ]
1
=
2
19
unit2 1