This document provides information about coordinate geometry, including finding the distance between two points, the midpoint and division of a line segment, area of polygons, and equations of straight lines. It gives formulas and examples for calculating the distance between points using the Pythagorean theorem, finding the midpoint and points dividing a line segment in a given ratio, and computing the area of triangles and quadrilaterals. It also explains how to determine the gradient, x-intercept, and y-intercept of a straight line and write the equation of a straight line in general and gradient forms. Exercises are provided to apply these concepts.

1. Additional Mathematics Module Form 4
Chapter 6- Coordinate Geometry SMK Agama Arau, Perlis
Page | 60
CHAPTER 6- COORDINATE GEOMETRY
6.1 DISTANCE BETWEEN TWO POINTS
y
y2 Q (x2, y2)
y2- y1
y1 P(x1, y1)
x2- x1
x
To find the distance or length of PQ, use the concept of Pythagoras’ Theorem.
2
12
2
12
2
)()( yyxxPQ −+−=
Therefore:
where D is distance.
Example 1:
The distance point A(6, 3t) and point B(12, -t) is 10 units. Find the possible values of t.
Solution:
2
4
6416
1636100
)4(6100
)3()612(10
2
2
2
22
22
±=
=
=
+=
−+=
−−+−=
t
t
t
t
t
tt
Example 2:
Point A(h, 2h) and point B(h -3 , 2h + 1) are two points which are equidistant from the origin. Find the
value of h.
Solution:
2222
)012()03()02()0( −++−−=−+− hhhh
2
12
2
12 )()( yyxxD −+−=
How to obtain the formula?
Square the both sides

2. Additional Mathematics Module Form 4
Chapter 6- Coordinate Geometry SMK Agama Arau, Perlis
Page | 61
5
102
10255
144964
)12()3()2(
22
2222
2222
=
=
+−=
++++−=+
++−=+
h
h
hhh
hhhhhh
hhhh
EXERCISE 6.1
1. Find the perimeter of triangle ABC with points A(2, 1), B(4, 5) and C(-2, 8).
2. The points (-3, -5) lies on the circumference of a circle with centre (2, 1). Calculate the radius of the
circle.
3. Given that the distance between points A(1,3) and B(7, k) is 10 units . Find the possible value of k.
4. Given point A( 1, 7) and B(p, 2) and the distance between the points A and B is 13 units. Find the value
of p.
6.2 DIVISION OF A LINE SEGMENT
Line segment is a line that has distant.
6.2.1 Mid-point
The formula to find mid-points is just the same as we have learned in Form Two that is:
Example:
Given C(2, 5) is the mid-point of the point B(h, 3) and point D(-4, k). Find the values of h and k.
Solution:
Use the formula,
8
44
2
4
2
=
=−
+−
=
h
h
h
7
103
2
3
5
=
=+
+
=
k
k
k
EXERCISE 6.2.1
1. The coordinates of A and B are (m, 5) and (6, n) respectively. Find the values of m and n if the mid-
point of the points is (4, 10).
2. The coordinates of M and N are (4, 2) and (6, 5) respectively. Find the mid-point of these two points.
3. The coordinates of X and Y are (-1, b) and (a, 7) respectively. Find the values of a and b if the mid-
point of the points is (1, 2).
)
2
,
2
(),( 2121 yyxx
yx
++
=

3. Additional Mathematics Module Form 4
Chapter 6- Coordinate Geometry SMK Agama Arau, Perlis
Page | 62
6.2.2 Point which divides a line segment in the ratio m: n
Q (x2, y2)
n
y2- y y2- y
X(x, y)
m y- y1
y- y1
P(x1, y1) x- x1 x2- x
If the line PQ moves downwards, it will reach the horizontal line.
From that we know that:
nm
mxnx
x
mxnxnmx
mxnxmxnx
mxmxnxnx
xxmxxn
n
m
xx
xx
+
+
=
+=+
+=+
−=−
−=−
=
−
−
21
21
21
21
21
2
1
)(
)()(
If the line PQ moves upwards, it will reach the vertical line
nm
myny
y
mynynmy
mynymyny
mymynyny
yymyyn
n
m
yy
yy
+
+
=
+=+
+=+
−=−
−=−
=
−
−
21
21
21
21
21
2
1
)(
)()(
Hence, the formula to find the point that divides the line segment in ratio m : n is
When nm = , the point will be the midpoint of the line segment.






+
+
+
+
=
nm
myny
nm
mxnx
yx 2121
,),(

4. Additional Mathematics Module Form 4
Chapter 6- Coordinate Geometry SMK Agama Arau, Perlis
Page | 63
Example:
Find the coordinates of point P that divides the straight line that joins E(-6, 10) and F(4, -5) in the ratio
2: 3
E(-6, 10)
Solution:
m: n = 2 : 3 2
P(x, y)
3
F(4, -5)
EXERCISE 6.2.2
1. Point R divides the line segment joining J(-1, -7) and Q(10 ,7) internally in the ration PR: RQ = 1: 3. Find
the coordinate of R.
2. The points P(t, 2t), Q(2a, b) and R(4a, 3b) are on a straight line. Q divides PR internally in the ratio 1: 4.
Show that ab 6= .
3. Given points A(k, 5), B(0, 3) and C(5, 4). Find the possible values of k if the length of AB is twice the
length of BC.
6.3 AREA OF POLYGONS
6.3.1 Area Of Triangle
C(x3, y3)
B(x2, y2)
A(x1, y1)
K L M
The area of ABC∆ = Area of trapezium ACLK + Area of trapezium BCLM - Area of trapezium ABMK
o
o
o






+
−+
+
+−
=
32
)5)(2()10)(3(
,
32
)4)(2()6)(3(
),( yxP






+
+
+
+
=
nm
myny
nm
mxnx
yx 2121
,),(





 −+−
=
5
1030
,
5
818
),( yxP
( )4,2−=
How to obtain the formula?

5. Additional Mathematics Module Form 4
Chapter 6- Coordinate Geometry SMK Agama Arau, Perlis
Page | 64
))((
2
1
))((
2
1
))((
2
1
122132321313 xxyyxxyyxxyy −+−−++−+=
[ ])()(
2
1
)(
2
1
)[
2
1
)([
2
1
312312133221
312312133221
211112223323322231111333
211112223323322231111333
yxyxyxyxyxyx
yxyxyxyxyxyx
yxyxyxyxyxyxyxyxyxyxyxyx
yxyxyxyxyxyxyxyxyxyxyxyx
++−++=
−−−++=
++−−−−++−−+=
−−+−−−++−−+=
Area of ∆ ABC =
1
1
3
3
2
2
1
1
2
1
y
x
y
x
y
x
y
x
=
[ ])()(
2
1
312312133221 yxyxyxyxyxyx ++−++=
The formula of the area of triangle is
6.3.2 Area of Quadrilateral
The formula of the area of quadrilateral is
( ) ( )4134231214433221
2
1
yxyxyxyxyxyxyxyx +++−+++=
Example 1:
Find the value of m if the point P(m, 2), Q(4, -3) and R(-2, 5) lie on a straight line.
Solution:
Method 1- Using Concept Area Of Triangle
0
25
2
3
4
22
1
=
−
−
mm
1
1
3
3
2
2
1
1
2
1
y
x
y
x
y
x
y
x
The area of a straight line is
zero.
1
1
4
4
3
3
2
2
1
1
2
1
y
x
y
x
y
x
y
x
y
x
Simplify

6. Additional Mathematics Module Form 4
Chapter 6- Coordinate Geometry SMK Agama Arau, Perlis
Page | 65
028
0514163
0)514(163
0)568()4203(
=+−
=−−+−
=+−+−
=++−−+−
m
mm
mm
mm
028
028
=+−
=+−
m
m
4
1
28
=
=
m
m
Method 2
Using the concept of gradient of straight line
6
8
)2(4
53
−
=
−−
−−
=
QR
QR
m
m
3
4
−=
4
1
14
15164
3
4
4
)3(2
=
=
−=−
−=
−
−−
=
m
m
m
m
mm QRPQ
Example 2:
Find the possible values of k if the area of triangle with vertices A (9, 2), B(4, 12) and C(k, 6) is 30 unit2
Solution:
30
2
9
612
4
2
9
2
1
=
k
601070
60)54128()224108(
=−
=++−++
k
kk
Point P, Q and R lie on the
same line, so QRPQ mm =

7. Additional Mathematics Module Form 4
Chapter 6- Coordinate Geometry SMK Agama Arau, Perlis
Page | 66
(i) 601070 −=− k
1
1010
=
=
k
k
(ii) 601070 −=− k
13
13010
=
=
k
k
Example 3:
Find the area of quadrilateral KLMN given K (1, 3), L(-1, 2) and M(-4, -3) and N(6, -9).
Solution:
3
1
9
6
3
4
2
1
3
1
2
1
−−
−−
=
2
5.48
97
2
1
3853
2
1
)38(59
2
1
)91883()183632(
2
1
unit=
=
+=
−−=
−−−−−+++=
EXERCISE 6.3
1. Find the possible values of p if the area of triangle with the vertices D(p, -p), E(1, 0) and F(-3, 6) is 10
unit2
.
2. Find the area of the rhombus PQRS if the coordinates of the points P, Q and R are (6, 4). (8, 7) and
(-6, 3) respectively.
3. The points G(4, -2), F(1, 1) and H(-2, p) lie on a straight line. Find the value of p.
4. Find the area of the triangle PQR if the coordinates of the vertices are:
(a) P(1, 3), Q (4, 2) and R (7, 0)
(b) P(-2, 5), Q (7, -2) and R (-3, 1)
(c) P(-1, -5),Q (1, -4) and R(-1, -3)
The modulus sign will always result a
positive number. If k1070 − would
result 60, there will be two values of k. If
the value of k is 1, the value in the
modulus is 60. If the value of k is 13, the
value in the modulus is -60 and at last
becomes 60 because the modulus sign
will always result a positive number.
1
1
4
4
3
3
2
2
1
1
2
1
y
x
y
x
y
x
y
x
y
x

8. Additional Mathematics Module Form 4
Chapter 6- Coordinate Geometry SMK Agama Arau, Perlis
Page | 67
6.4 EQUATION OF A STRAIGHT LINE
6.4.1 x-intercept and y-Intercept of a straight line
y
b
x
0 a
1. The line intersects with the x-axis at a and the line intersects with the y-axis at b
2. a is called x-intercept and b is called y-intercept.
EXERCISE 6.4.1
1. Determine the x-intercept for the following straight lines:
(a) 1
32
=+
yx
(b) 0943 =−− yx
(c) 66 −= xy
2. Determine the y-intercept for the following straight lines:
(a) 1
51
=+
yx
(b) 01475 =−− yx
(c) 653 −= xy
6.4.2 The Gradient of a Straight Line
(i) (ii) (iii) (iv)
The gradient is positive the gradient is negative the gradient is undefined the gradient is zero

9. Additional Mathematics Module Form 4
Chapter 6- Coordinate Geometry SMK Agama Arau, Perlis
Page | 68
Finding the gradient (m) of a straight line
1-
Example:
y
A(0,5)
x
0 B(0, 4)
12
12
xx
yy
m
−
−
=
40
05
−
−
=ABm
4
5
−=
2-
y
(0, b)
x
0 (a, 0)
a
b
m
−
−
=
0
0
12
12
xx
yy
m
−
−
=

10. Additional Mathematics Module Form 4
Chapter 6- Coordinate Geometry SMK Agama Arau, Perlis
Page | 69
a
b
−=
Hence,
Example:
y
A(0,5)
x
0 B(0, 4)
erceptx
ercepty
m
int
int
−
−
=
4
5
−=ABm
3-
Example: y
A(0,5)
5
x
0 4 B(0, 4)
θ
erceptx
ercepty
m
int
int
−
−
−=
θtan=m

11. Additional Mathematics Module Form 4
Chapter 6- Coordinate Geometry SMK Agama Arau, Perlis
Page | 70
θtan=ABm
4
5
−=ABm
EXERCISE 6.4.2
Find the gradient of the following points.
(a) A (2, 3) and B (4, 5)
(b) M (-3, 1) and N (4, -2)
(c) P (2, 3) and Q (4, 3)
(d) C (2, 5) and D (2, 8)
(e) F(0, 6) and G( 3, 0)
6.4.3 The Equation of a Straight Line
1-General Form
The equation of general form is 0=++ cbyax
Example:
Given the equation of a straight line is 542 += xy . Change the equation into the general form.
Solution:
542 += xy
0524 =+− yx
2-Gradient Form
The equation of gradient form is cmxy += where m is the gradient and c is y-intercept
Example:
Given the equation of a straight line is 542 += xy . Determine the gradient and the y-intercept of
the straight line.
Solution:
542 += xy
2
5
2 += xy
Hence, the gradient of the straight line is 2 while and the y-intercept of the straight line is
2
5
.

12. Additional Mathematics Module Form 4
Chapter 6- Coordinate Geometry SMK Agama Arau, Perlis
Page | 71
3-Intercept form
The equation of intercept form is 1=+
b
y
a
x
where a is x-intercept and b is y-intercept.
Example:
Given the equation of a straight line is 632 += xy . Convert the equation into the intercept form.
Hence, state the x-intercept and y-intercept of the straight line.
Solution:
632 += xy
1
32
1
32
1
6
2
6
3
623
=+
−
=+
−
=+
−
=+−
yx
yx
yx
yx
Hence, the x-intercept of the straight line is -2 and y-intercept of the straight line is 3.
Example:
Find the equation of the straight line which has a gradient of -3 and passes through the mid-point of the
line joining A (1, 4) and B(7, -2).
Solution:
Mid-point of AB 




 −++
=
2
)2(4
,
2
17
( )1,4=
The equation of the straight line which has a gradient of -3 and passes through (4, -1) is
xy
x
y
3121
3
4
1
−=−
−=
−
−
0133 =−+ yx or 133 +−= xy
EXERCISE 6.4.3
1. Write each of the following equations to intercept form. Hence, state the gradient of the straight line.
(a) 042 =−+ yx
(b) 63 =− yx
(c) 234 =+ yx
(d) 132 =+ yx
2. Find the equation of the straight line which has a gradient of -2 and passes through point B (7, -2).
Use the general point (x, y) and
specific point (4, -1) to find the
gradient of the line and it is equal
to the given gradient.

13. Additional Mathematics Module Form 4
Chapter 6- Coordinate Geometry SMK Agama Arau, Perlis
Page | 72
6.4.4 The Point of intersection of two Straight lines
1. When two lines intersect, the point of intersection lies on both lines.
2. This means the coordinates of the point satisfy both the equations of the lines.
3. Therefore, we need to solve the equation simultaneously in order to determine the point of
intersection.
Example 1:
The straight line which has a gradient of 2 and passes through the point (4, -1) intersects with the
straight line 6−=+ yx at the point P. Find the coordinates of the point P.
Solution:
First of all, we have to find the equation of the straight line.
92
821
2
4
)1(
=−
−=+
=
−
−−
yx
xy
x
y
6−=+ yx
+ ,
1
33
=
=
x
x
Substitute 1=x into ,
7
9)1(2
−=
=−
y
y
Hence the coordinates of P is (1, -7)
Example 2:
The straight line 04 =−+ yx and 01132 =−+ yx intersect at point A. Find the equation of the
straight line which passes through the point A and point B (5, 2).
Solution:
0822
04
=−+
=−+
yx
yx
01132 =−+ yx
- ,
3
03
=
=−
y
y
1
2
1 2
1
1
2
2 1

14. Additional Mathematics Module Form 4
Chapter 6- Coordinate Geometry SMK Agama Arau, Perlis
Page | 73
Substitute 3=y into ,
1
22
08)3(22
=
=
=−+
x
x
x
Hence the coordinates of A is (1, 3)
15
32
−
−
=ABm
4
1
−=ABm
The equation of the straight line that passes through point A and B is
xy
x
y
−=−
−=
−
−
1124
4
1
1
3
0134 =−+ yx or 134 =+ yx or
4
13
4
1
+−= xy
EXERCISE 6.4.4
1. Find the coordinates of the point of intersection of the line 01134 =−+ yx and 01762 =+− yx .
2. Find the points of intersection of the following pairs of straight lines.
(a) 025 =−− yx (b) 42 += xy
052 =+− yx 5+= xy
3. Find the equation of that is parallel to the line 52 += xy and passing through the point of
intersection of lines 092 =−− yx and 22 =+ yx .
6.5 PARALLEL AND PERPENDICULAR LINES
Parallel Lines
1. When two lines are parallel, they have the same gradient.
B Q
A P
2. If line AB and line PQ are parallel, so PQAB mm = .
1
We can write the equation in any
form. Either the general form or
intercept form or gradient form.

15. Additional Mathematics Module Form 4
Chapter 6- Coordinate Geometry SMK Agama Arau, Perlis
Page | 74
Example:
The straight line AB passes through the point (6, 3) is parallel to the straight line PQ. Given point P (0, 2)
and point Q(4, 0). Find the equation of the straight line AB.
Solution:
First of all, we have to find the gradient of straight line PQ
4
2
−=PQm
2
1
−=
PQAB mm =
2
1
−=ABm
The equation of the straight line that passes through point (6, 3) is
xy
x
y
−=−
−=
−
−
662
2
1
6
3
0122 =−+ yx or 122 =+ yx or 6
2
1
+−= xy
Perpendicular Lines
y
x
1. Given that line AB and BC are perpendicular to each other.
2. We already know θtan=m .
θtan1 =m αtan2 =m
AB
BC
=
BC
AB
−=
B
C
A
θ α
1m
2m
We can write the equation in any
form. Either the general form or
intercept form or gradient form.

16. Additional Mathematics Module Form 4
Chapter 6- Coordinate Geometry SMK Agama Arau, Perlis
Page | 75
3. When
BC
AB
AB
BC
mm −×=× 21
Hence, if two lines are perpendicular to each other, then the product of their gradient is 1− .
Example:
Given the straight line 9−= txy and 32 += xy is perpendicular to each other. Find the value of t.
Solution:
32 += xy
21 =m
9−= txy
tm =2
121 −=× mm
2
1
12
12
−=
−=
−=×
t
t
t
EXERCISE 6.5
1. The equation of the straight line PQ is 0786 =+− xx . Each of the following straight line is parallel to
PQ. Find the value of t in each case.
(a) 064 =−+ ytx
(b) 8
2
+= x
t
y
(c) 012 =−− tyx
2. Find the equation of the straight line which passes through point B (2, -5) and perpendicular to the
straight line 13 +−= xy .
3. PQRS is a rhombus with P (0, 5) and the equation of QS is 12 += xy . Find the equation of diagonal of
PR.
4. Find the value of h if the straight line 02 =+− hxy is perpendicular to the straight line
035 =++ xy .
5. Given that the equation of the line PQ is 1532 += xy and point Q lies on the y-axis. Point R is (4, 1)
lies on line QR. Find the equation of QR if the line PQ and QR are perpendicular to each of other.
Use the concept
121 −=× mm

17. Additional Mathematics Module Form 4
Chapter 6- Coordinate Geometry SMK Agama Arau, Perlis
Page | 76
6.6 LOCUS OF A MOVING POINT
Locus represents the path followed by a moving point with the reference to one or more fixed points,
satisfying certain conditions.
6.6.1 Equation of Locus
Equidistant from a fixed point
y
P (x, y)
3 unit
A (1, 1)
x
The equation of locus is
0722
91212
9)1()1(
3)1()1(
22
22
22
22
=−−−+
=+−++−
=−+−
=−+−
yxyx
yyxx
yx
yx
Equidistant from two fixed points
Find the equation of the locus of a moving point P such that its distance from the point A (1, 2) and point
B (3, 4) are equal.
B (3, 4)
A (1, 2)
Equation of locus is actually
involving the distance between
two points. So we have to use
the formula of the distance
between two points to find the
equation of locus. There is no
specific formula to find the
formula to find the equation of
locus.
In this case, P is the moving point
such that its distance is always 3
unit from point A.
05
02044
2586542
168964412
)4()3()2()1(
)4()3()2()1(
2222
2222
2222
2222
=−+
=−+
+−−+=+−−+
+−++−=+−++−
−+−=−+−
−+−=−+−
=
yx
yx
yxyxyxyx
yyxxyyxx
yxyx
yxyx
BPAP
locus
),( yxP
Square the both sides
Square the both sides

18. Additional Mathematics Module Form 4
Chapter 6- Coordinate Geometry SMK Agama Arau, Perlis
Page | 77
Constant ratio between two fixed points
Find the equation of locus of a moving point R such that its distance from the point E (4, 3) and the point
F (1, 5) is in the ration 3: 1.
0209841088
961682259099189
96168)251012(9
)3()4()5()1(3
3
1
3
22
2222
2222
2222
=+−−+
+−++−=+−++−
+−++−=+−++−
−+−=−+−
=
=
yxyx
yyxxyyxx
yyxxyyxx
yxyx
RERF
RF
RE
EXERCISE 6.6
1. Given the point A (0, 3) and the point B (1, 4).Find the equation of locus of a moving point Q such that
AQ= 2QB.
2. Given A (5, -2) and B (2, 1) are two fixed points. Point Q moves such that the ratio of AQ: QB = 2: 1.
Show that the equation of the locus of point Q is 034222
=−−−+ yxyx .
3. P is a moving point such that its distances from the points A(2, 5) and B(0, 3) is in the ratio of 2: 1. Find
the equation of locus P.
4. N is a locus which moves in such a way that NP=NQ. Given that P and Q are coordinates (-3, 6) and
(6,- 4) respectively, find the equation of locus N.
5. Show that the equation of the locus of a point that moves in such way that is distance from a fixed
point (3, -1) is 6 units, is by 0262622
=−+−+ yxyx .
CHAPTER REVIEW EXERCISE
1. Given the equation of straight lines AB and CD are 1
6
=+
k
yx
and 0432 =−+ yx respectively, find
the value of k if AB is perpendicular to CD.
2. The coordinates of the point A and B are (-2, 3) and (7, -3) respectively. Find
(a) the coordinates of C given that AB: BC = 1: 2.
(b) the equation of the straight line that passes through B and is perpendicular to AB.
3. ABCD is a parallelogram with coordinates A (-2, 3), B (3,4), C (2, -1) and D (h, k).
(a) Find the value of h and k.
(b) Find the equations of the diagonals AC and BD.
(c) State the angle between the diagonals AC and BD.
(d) Find the area of the parallelogram ABCD.
Square the both sides

19. Additional Mathematics Module Form 4
Chapter 6- Coordinate Geometry SMK Agama Arau, Perlis
Page | 78
4. P, Q and R are three points on a straight line. The coordinates of P and R are (-2, 3) and (3, 5)
respectively. Point Q lies on the y-axis. Find
(a) the ratio PQ: QR
(b) the coordinates of point Q
5. H is a point which moves such that its distance from point P (1, -2) and Q (-3, 4) is always equal. Show
that the equation of the locus H is given by the equation 0532 =+− yx .
6. Find the equation of straight line that passes through point P( 1 -2) and parallel to 4x – 2y = 8.
7. In diagram below, °=∠ 90PRS .
Find
(a) the coordinates of R
(b) the ratio of PQ: QR
(c) the equation of RS.
8. Diagram below shows a triangle BCD. The point A lies on the straight line BD.
Find
(a) the value of k
(b) the equation of CD, giving your answer in general form.
)9,6(−P
S
Q
R
3
x
y
0
)7,1(−C
)8,13(A
)2,5(B
),17( kD
x
0
y