Chapter 7

Coordinate Geometry

7.1 Midpoint of the Line Joining Two Points

7.2 Areas of Triangles and Quadrilaterals

7.3 Parallel and Non-Parallel Lines

7.4 Perpendicular Lines

Coordinate Geometry

Objectives

7.1 Midpoint of the Line Joining Two Points

In this lesson, you will learn how to find the midpoint of a line segment and apply it to solve problems.

A line AB joins points (x1, y1) and (x2, y2).

M (x, y) is the midpoint of AB.

y

x

A ( x1 , y1)

B ( x2 , y2)y

x

A ( x1 , y1)

B ( x2 , y2)

M ( x , y)

y

x

A ( x1 , y1)

B ( x2 , y2)

M ( x , y)

C ( x2 , y1)

y

x

A ( x1 , y1)

B ( x2 , y2)

M ( x , y)

C ( x2 , y1)D ( x , y1)

E ( x2 , y)

y

x

A ( x1 , y1)

B ( x2 , y2)

M ( x , y)

C ( x2 , y1)D ( x , y1)

E ( x2 , y)

Construct a right angled triangle ABC.

Construct the midpoints D and E of the line segments AC and BC. Take the mean of the coordinates at the endpoints.

D is and E is1 21,

2

x xy

1 21, 2

y yx

M is the point 1 2 1 2,2 2

x x y y

Take the x-coordinate of D and the y-coordinate of

E.

Coordinate Geometry

y

x

S ( a , b)

Q ( 9 , 6)

R ( – 2 , 4)

P ( 4 , – 4)

O

4 2 4 4, 1,0

2 2M

P, Q, R and S are coordinates of a parallelogram and M is the midpoint of PR. Find the coordinates of M and S and show that PQRS is a rhombus.

y

x

S ( a , b)

Q ( 9 , 6)

R ( – 2 , 4)

P ( 4 , – 4)

O

M

y

x

S ( a , b)

Q ( 9 , 6)

R ( – 2 , 4)

P ( 4 , – 4)

OM ( 1 , 0)

y

x

S ( a , b)

Q ( 9 , 6)

R ( – 2 , 4)

P ( 4 , – 4)

OM ( 1 , 0) 9 6

, 1,02 2

a bM

7, 6a b = 7, 6S

M is also the midpoint of QS.

229 4 6 4 125PQ

2 29 2 6 4 125QR

the parallelogram is a rhombusPQ QR PQRS

y

x

S ( – 7 , – 6)

Q ( 9 , 6)

R ( – 2 , 4)

P ( 4 , – 4)

OM ( 1 , 0)

Coordinate Geometry

Example 4

y

x

B ( 3 , 2)

A ( – 1 , 6)

C ( – 5 , – 4)

1 3 6 2, 1,4

2 2D

3 points have coordinates A(–1, 6), B(3, 2) and C(–5, –4). Given that D and E are the midpoints of AB and AC respectively, calculate the midpoint and length of DE.

1 5 6 4, 3,1

2 2E

Let M be the midpoint of DE.

2 23 1 1 4 5DE

y

x

B ( 3 , 2)

A ( – 1 , 6)

C ( – 5 , – 4)

D

E

y

x

B ( 3 , 2)

A ( – 1 , 6)

C ( – 5 , – 4)

D ( 1 , 4)

E

y

x

B ( 3 , 2)

A ( – 1 , 6)

C ( – 5 , – 4)

D ( 1 , 4)

E ( – 3 , 1)

12

1 3 4 1, 1,2

2 2M

y

x

B ( 3 , 2)

A ( – 1 , 6)

C ( – 5 , – 4)

D ( 1 , 4)

E ( – 3 , 1) M

y

x

B ( 3 , 2)

A ( – 1 , 6)

C ( – 5 , – 4)

D ( 1 , 4)

E ( – 3 , 1)

M ( – 1 , 2 12 )

Coordinate GeometryExercise 7.1, qn 3

y

x

B ( p , – 2)

D ( 3 , r )

A ( 2 , 0)

C ( – 1 , 1)

1 12 2

2 1 0 1, ,

2 2M

1 12 2

23, ,

2 2

rpM

2, 3p r

Let M be the midpoint of AC.

If A(2, 0), B(p, –2), C(–1, 1) and D(3, r) are the vertices of a parallelogram ABCD, calculate the values of p and r.

y

x

B ( p , – 2)

D ( 3 , r )

A ( 2 , 0)

C ( – 1 , 1)M

y

x

B ( p , – 2)

D ( 3 , r )

A ( 2 , 0)

C ( – 1 , 1)M ( 1

2 , 1

2 )

M is also the midpoint of BD.

y

x

B ( p , – 2)

D ( 3 , r )

A ( 2 , 0)

C ( – 1 , 1)M ( 1

2 , 1

2 )

y

x

B ( – 2 , – 2)

D ( 3 , 3 )

A ( 2 , 0)

C ( – 1 , 1)M ( 1

2 , 1

2 )

Coordinate Geometry

Exercise 7.1, qn 4

7.2 Areas of Triangles and Quadrilaterals

In this lesson, you will learn how to find the areas of rectilinear figures given their vertices.

Coordinate Geometry

Objectives

ABC is a triangle. We will find its

area.

Construct points D and E so that ADEC is a

trapezium.

y

x

A ( – 2 , – 1)

B ( 2 , – 3)

C ( 4 , 3)

y

x

A ( – 2 , – 1)

B ( 2 , – 3)

C ( 4 , 3)

D ( – 2 , – 3) E ( 4 , – 3)

Area of Area of trapezium Area of Area of ABC ADEC ADB BEC

1 1 12 6 6 4 2 2 6

2 2 2

24 4 6 14 square units

Coordinate Geometry

Area of Triangles

ABC is a triangle. The vertices are arranged in

an anticlockwise direction. We will find

its area.

Construct points D, E and F on the x-axis as shown.

y

x

C (x3, y3 )

A (x1, y1 )

B (x2, y2 )

Area of ABC

1 1 12 2 2BE AF EF CD BE DE CD AF DF

11 2 2 3 3 1 2 1 3 2 1 32 x y x y x y x y x y x y

y

x

C (x3, y3 )

A (x1, y1 )

B (x2, y2 )

D (x3, 0 ) E (x2, 0 ) F (x1, 0 )

1 1 12 1 1 2 3 2 2 3 3 1 1 32 2 2y y x x y y x x y y x x

y

x

C (x3, y3 )

A (x1, y1 )

B (x2, y2 )

D (x3, 0 ) E (x2, 0 ) F (x1, 0 )

11 2 1 1 2 2 2 1 2 3 2 2 3 2 3 3 1 3 1 1 3 1 3 32 x y x y x y x y x y x y x y x y x y x y x y x y

Area of Area of Area of ABEF BCDE ACDF

Coordinate Geometry

y

x

C (x3, y3 )

A (x1, y1 )

B (x2, y2 )

1 2 2 3 3 1 2 1 3 2 1 3

1Area of

2ABC x y x y x y x y x y x y

1 2 3 1

1 2 3 1

1

2

x x x x

y y y y

1 2 3 1

1 2 3 1

x x x x

y y y y

1 2x y1 2 2 3 3 1x y x y x y 1 2 2 3x y x y1 2 2 3 3 1 2 1 3 2x y x y x y x y x y 1 2 2 3 3 1 2 1x y x y x y x y 1 2 2 3 3 1 2 1 3 2 1 3x y x y x y x y x y x y

From the previous slide, we know that

Definition

Coordinate Geometry

y

x

A ( – 2 , – 1)

B ( 2 , – 3)

C ( 4 , 3)

Area of 2 2 4 21

1 3 3 12ABC

16 6 4 2 12 6

2

14 square units

Find the area of a triangle with vertices A(–2, –1), B(2, –3) and C(4, 3).

The vertices A, B and C follow an anticlockwise direction.

Coordinate Geometry

Example 5

y

x

B ( x2 , y2 )

D ( x4 , y4 )

A ( x1 , y1 )

C ( x3 , y3 )

Area of ABCD ABC ACD

1 2 2 3 3 4 4 1 2 1 3 2 4 3 1 4

1

2x y x y x y x y x y x y x y x y

Find the area of a quadrilateral with vertices A(x1, y1 ), B(x2, y2 ), C(x3, y3 ) and D(x4, y4 ), following an anticlockwise direction.

y

x

B ( x2 , y2 )

D ( x4 , y4 )

A ( x1 , y1 )

C ( x3 , y3 )

Split the quadrilateral into two triangles.

1 2 3 1 1 3 4 1

1 2 3 1 1 3 4 1

1 1

2 2

x x x x x x x x

y y y y y y y y

1 2 2 3 3 1 2 1 3 2 1 3

1Area of

2ABCD x y x y x y x y x y x y

1 3 3 4 4 1 3 1 4 3 1 4

1

2x y x y x y x y x y x y

Coordinate Geometry

Area of Quadrilaterals

y

x

B ( x2 , y2 )

D ( x4 , y4 )

A ( x1 , y1 )

C ( x3 , y3 )

1 2 2 3 3 4 4 1 2 1 3 2 4 3 1 4

1

2x y x y x y x y x y x y x y x y

The area of a quadrilateral with vertices A(x1, y1 ), B(x2, y2 ), C(x3, y3 ) and D(x4, y4 ), following an anticlockwise direction.

The method for finding the area of quadrilaterals is very similar to that of triangles.1 2 3 4 1

1 2 3 4 1

1

2

x x x x x

y y y y y

Coordinate Geometry

y

x

S ( 4 , 0 )

Q ( – 4 , 3 )

R ( 1 , – 2 )

P ( 1 , 4 )

13 8 0 16 16 3 8 0

2

Find the area of a quadrilateral with vertices P(1, 4 ), Q(–4, 3), R(1, –2) and S(4, 0), following an anticlockwise direction.

Area1 4 1 4 11

4 3 2 0 42

24 square units

Coordinate Geometry

Exercise 7.2, qn 2(b)

7.3 Parallel and Non-Parallel Lines

In this lesson, you will learn how to apply the conditions for the gradients of parallel lines to solve problems.

Coordinate Geometry

Objectives

y

xO

y = m1x + c1

1

Consider the straight line with equation y = m1 x + c1 that makes an angle of θ1

with the positive x-axis.

Translate the line parallel to the x-axis.

y

xO

y = m1x + c1 y = m2x + c2

1 2

y

xO

y = m1x + c1 y = m2x + c2

1 2

The new line has equation y = m2 x + c2 and makes an angle of θ2

with the positive x-axis.

The lines are parallel to each other.

The lines make the same angle with the x-axis.

The lines have the same gradient.

θ1 = θ2

m1 = m2

Coordinate Geometry

y

xO

A

B

C

D

x + y = 2

2y = x + 10

At , 0.A y

The diagram shows a parallelogram ABCD with A and C on the x-axis and y-axis respectively. The equation of AB is x + y = 2 and the equation of BC is 2y = x + 10.

Since 2, 2.x y x

is 2,0A

At , 22 10

B x yy x

Solving 2, 4.x y

is 2,4 .B

At , 0.C x

Since 2 10, 5.y x y

is 0,5 .C

(a) Find the coordinates of A, B and C.

Coordinate Geometry

Example 7(a)

y

xO

A

B

C

D

x + y = 2

2y = x + 10

0Equation of is

1

2 2

yAD

x

The diagram shows a parallelogram ABCD with A and C on the x-axis and y-axis respectively. The equation of AB is x + y = 2 and the equation of BC is 2y = x + 10.

2 2y x

(b) Find the equations of AD and CD.

AD is parallel to BC (2y = x + 10).

Gradient of AD = gradient of BC = 0.5

Since A is (2, 0)

5Equation of is 1

0

yAD

x

5x y

CD is parallel to AB (x + y = 2 ).

Gradient of CD = gradient of AB = –1

Since C is (0, 5)

y

xO

A

B

C

D

x + y = 2

2y = x + 10

y

xO

A

B

C

D

x + y = 2

2y = x + 10

(2, 0)

y

xO

A

B

C

D

x + y = 2

2y = x + 10

(2, 0)

2y = x – 2

y

xO

A

B

C

D

x + y = 2

2y = x + 10

2y = x – 2

y

xO

A

B

C

D

x + y = 2

2y = x + 10

2y = x – 2

(0, 5)

y

xO

A

B

C

D

x + y = 2

2y = x + 10

2y = x – 2

x + y = 5

Coordinate Geometry

Example 7(b)

y

xO

2x + 3y – 3 = 0

( – 2, 3)

2 3 3 0x y

Find the equation of the line which passes through the point (–2, 3) and is parallel to the line 2x + 3y – 3 = 0.

3 2 3y x 23 1y x

23The gradient of the line is .

23The line through 2,3 with gradient is

Rearrange in the form y = mx + c.

m in y = mx + c is the gradient

of the line.

3 2

2 3

y

x

3 3 2 2y x 3 9 2 4y x

The equation of the line is 2 3 5 0.x y

y

xO

2x + 3y – 3 = 0

( – 2, 3)

2x + 3y – 5 = 0

Coordinate Geometry

Exercise 7.3, qn 2(b)

Coordinate Geometry

7.4 Perpendicular Lines

In this lesson, you will learn how to apply the conditions for the gradients of perpendicular lines to solve problems.

Objectives

y

xO

y = m1x + c1

1

Consider the straight line with equation y = m1 x + c1 that makes an

angle of θ1 with the positive x-axis.

Rotate the line clockwise through 90°.

y

xO

y = m1x + c1

1

y = m2x + c2

2

The new line has equation y = m2 x + c2

and makes an angle of θ2 with the

negative x-axis.

y

xO

y = m1x + c1

1

y = m2x + c2

2

A

B CD

1 1tanAD AC

mBD AB

2 2tanAD AB

mDC AC

1 2 1AC AB

m mAB AC

21

1m

m Applies to any two

perpendicular lines.

Coordinate Geometry

y

xO

B(4, 15)

A( – 2, 3)3y = x + 1 The midpoint of is

2 4 3 15, 1,9

2 2AB M

Two points have coordinates A(–2, 3) and B(4, 15). Find the equation of the perpendicular bisector of AB. Hence calculate the coordinates of the point P on the line 3y = x + 1 if P is equidistant from A and B.

Gradient of

15 32

4 2AB

12Gradient of perpendicular bisector =

2 193 1

y xy x

5 20 4y y

The midpoint of is .AB M

y

xO

B(4, 15)

A( – 2, 3)3y = x + 1

M

y

xO

B(4, 15)

A( – 2, 3)3y = x + 1

M (1, 9)

y

xO

B(4, 15)

A( – 2, 3)3y = x + 1

M (1, 9)

2y = – x + 19

12

Equation of perpendicular bisector is 9 1y x

2 19y x 12 1 11x x

Find P.

11,4P

y

xO

B(4, 15)

A( – 2, 3)3y = x + 1

M (1, 9)

2y = – x + 19

P

y

xO

B(4, 15)

A( – 2, 3)3y = x + 1

M (1, 9)

2y = – x + 19

P(11, 4 )

Solve simultaneous equations

Substitute for y

Adding the

equations

Coordinate Geometry

Example 14

y

xO

B(3, 6)

A(5, 2)P

Q

The points A and B have coordinates (5, 2) and (3, 6) respectively. P and Q are points on the x-axis and y-axis and both P and Q are equidistant from A and B.

Gradient of 6 2

23 5

AB

12Gradient of perpendicular bisector = At 0, 2 4P y y x

12

Equation of perpendicular bisector is 4 4y x

2 4y x

4x 4,0P

(a) Find the equation of the perpendicular bisector of AB.

(b) Find the coordinates of P and Q.

The midpoint of is 3 5 6 2

, 4,42 2

AB M

The midpoint of is .AB M

y

xO

B(3, 6)

A(5, 2)P

Q

M

y

xO

B(3, 6)

A(5, 2)P

Q

M(4, 4)

y

xO

B(3, 6)

A(5, 2)P

Q

M(4, 4)2y = x + 4

At 0, 2 4Q x y x

2y 0,2Q

y

xO

B(3, 6)

A(5, 2)P

Q

M(4, 4)2y = x + 4

( – 4, 0)

y

xO

B(3, 6)

A(5, 2)P

Q

M(4, 4)2y = x + 4

( – 4, 0)

(0, 2)

Coordinate Geometry

Exercise 7.4, qn 9