Term 3 : Unit 2Coordinate Geometry

Name : _____________ ( ) Class : ________ Date : ________

2.1 Midpoint of the Line Joining Two Points

2.2 Areas of Triangles and Quadrilaterals

2.3 Parallel and Non-Parallel Lines

2.4 Perpendicular Lines

2.5 Circles

Coordinate Geometry

Objectives

2.1 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 1

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 GeometryExample 2

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

Example 3

2.2 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 4

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

Example 5

3.3 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 6(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 6(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

Example 7

Coordinate Geometry

2.4 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 8

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

Example 9

Curves and Circles

In this lesson, you will learn to • recognise the equation of a circle, • find the centre and radius of a circle, • find the intersection of a circle and a straight line.

2.5 Circles

Objectives

y

x

C(a, b)

P(x, y)

O

y

x

C(a, b)

P(x, y)

O

Consider the point C(a, b) and a point P(x, y).

The equation of the circle is

2 2PC x a y b

The locus of P as the line rotates around C is a circle of radius r.

The distance between C and P is

2 2 2x a y b r

Curves and Circles

y

x

C(a, b)

P(x, y)

O

r

The circle with centre C(a, b) and radius r has equation

2 2 2 2 22 2x ax a y by b r

The general form of the equation is

2 2 2x a y b r

2 2 2 2 22 2 0x y ax by a b r

2 2 2 2 0x y gx fy c

2 2 2,g a f b c a b r centre ,C g f

2 2radius f g c

Curves and Circles

2 2 23 4 4x y

Find the equation of the circle whose centre is C(–3, 4) and which touches the x–axis.

The radius of the circle is 4 units.

y

x

C( – 3, 4)

O

y

x

C( – 3, 4)

4

O

The equation of the circle is:

2 26 9 8 16 16x x y y

2 2 6 8 9 0x y x y The radius is the y-coordinate of C.

Curves and Circles

Example 10

2 12y x

Find the coordinates of the points of intersection of the line 2y + x = 12 with the circle x2 + y2 – 6x – 4y – 12 = 0.

x5 10

y

5

2y + x = 12

x2

+ y2

– 6x – 4y – 12 = 0

The points are

2 2 6 4 12 0x y x y

2 2144 48 4 72 12 4 12 0y y y y y

12 2x y

2 212 2 6 12 2 4 12 0y y y y

25 40 60 0y y 5 2 6 0y y

2 or 6y y

8 or 0x x 8,2 and 0,6

Substitute for x into the circle

equation.

Using x = 12 – 2y.

Curves and Circles

Example 11

2 2 22 0 3x y

(a) Give the equation of the circle with centre C(–2, 0) and radius r = 3.

2 2 4 5 0x y x

2 2 2 2 0x y gx fy c

2 24 4 9x x y

2 2, 1g g

1c

1,3

The equation is

(b) Find the coordinates of the centre and the radius of the circle

x2 + y2 – 2x – 6y + 1 = 0.

centre ,C g f

2 2radius 3 1 1 2 6, 3f f

3

Curves and Circles

Example 12