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Coordinate Geometry. 2002 Paper 2 Question 2. 3x - 2y = 12 ---- (2). To find a point of intersection of 2 lines solve by doing a simultaneous equation. 4x + y = 5 ----- (1). Equalise the y terms. 3x - 2y = 12 ----- (2). 8x + 2y = 10 ----. (1) x 2. Add up.

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coordinate geometry

Coordinate Geometry

2002 Paper 2 Question 2

slide3
3x - 2y = 12 ---- (2)

To find a point of intersection of 2 lines solve by doing a simultaneous equation

4x + y = 5 ----- (1)

Equalise the y terms

3x - 2y = 12 ----- (2)

8x + 2y = 10 ----

(1) x 2

Add up

11x = 22 ----- (1) + (2)

x = 2

slide4
Substitute 2 for x in (1)

Check the results

4x + y =

4(2) + y = 5

5

4(2) + (-3) =

8 + y = 5

8 - 3 =

5

y = 5 - 8

3x – 2y =

y = - 3

3(2) – 2(- 3) =

12

x = 2 y = - 3

6 + 6 =

12

slide5
Testing a(0, 8)

Testing b(-10, 0)

4x – 5y =

- 40

4x – 5y =

- 40

4(0) – 5(8) =

4(-10) – 5(0) =

- 40

- 40

a(0, 8) is on L

b(-10, 0) is on L

slide6
x1 = 0

y1 = 8

x2 = -10

y2 = 0

y2 – y1

Slope of L =

x2 – x1

(0) – (8)

=

(-10) – (0)

-8

=

-10

4

=

5

slide7
4

-5

Slope of L =

Slope of K =

5

4

-5

x1 = -10

y1 = 0

m =

4

-5

(x – (-10))

y – 0 =

4

4.

-5

4y =

(x – (-10))

4

Equation of Line: y – y1 = m(x – x1)

K  L

Multiply by LCD = 4

b(-10, 0)  K

= -5(x + 10)

= -5x - 50

slide8
4y = - 5x - 50

5x + 4y + 50 = 0 Equation of K

slide9
- 25

=

2

- 50

y =

4

5x + 4y + 50 = 0 --- Equation of K

K intersects y-axis when x = 0

5(0) + 4y + 50 = 0

4y = - 50

= - 12½

C is (0, - 12½)

slide10
a(0, 8)

8

b(-10, 0)

12.5

c(0, -12.5)

Area of abc = ½ b h

= ½ (20.5)(10)

= 102.5

10

20.5

d

Area of abcd = 2(102.5)

= 205cm2

slide11
a(0, 8)

b(-10, 0)

c(0, -12.5)

bc maps b(-10, 0)  c(0, -12.5)

The translation that

maps b onto c

also

maps a onto d

d

a(0,8)  d(

10,

- 4.5)

coordinate geometry1

Coordinate Geometry

2001 Paper 2 Question 2

slide14
3x + 2y + 7 = 0

(t, 2t) is on the line

 3(t) + 2(2t) + 7 = 0

 3t + 4t + 7 = 0

 7t = - 7

 t = - 1

slide15
x1 = 4

y1 = 2

x2 = 0

y2 = 4

y2 – y1

Slope of ac =

x2 – x1

(4) – (2)

=

(0) – (4)

2

=

- 4

a(4, 2) b(-2, 0) c(0, 4)

slide16
x1 = - 2

y1 = 0

x2 = 0

y2 = 4

y2 – y1

Slope of bc =

x2 – x1

(4) – (0)

=

(0) – (-2)

4

=

2

a(4, 2) b(- 2, 0) c(0, 4)

= 2

slide17
Slope of ac

1

Product of slopes = 2 x

-

2

1

= -

2

Slope of bc = 2

= -1

 ac  bc

slide18
a(4, 2), b(-2, 0) c = (0, 4)

x1 = 4 x2 = 0

y1 = 2 y2 = 4

slide19
a(4, 2), b(-2, 0) c = (0, 4)

x1 = -2 x2 = 0

y1 = 0 y2 = 4

 |ac| = |bc|

slide20
c(0, 4)

a(4, 2)

b(-2, 0)

b(-2, 0)

o

Using the translation b(-2, 0)  o(0, 0)

a(4, 2)  p(6, 2)

c(0, 4)  p(2, 4)

slide21
q(2, 4)

c(0, 4)

a(4, 2)

p(6, 2)

b(-2, 0)

o

Area of abc = Area of opq = ½|x1y2 – x2y1|

x1 = 6 x2 = 2

y1 = 2 y2 = 4

slide22
x1 = 6 x2 = 2

y1 = 2 y2 = 4

Area of abc = Area of opq = ½|x1y2 – x2y1|

= ½|(6)(4) – (2)(2)|

= ½|24 – 4|

= ½(20)

= 10 cm2

slide23
g

h

c(0, 4)

b(-2, 0)

a(4, 2)

The translation that maps b(-2, 0)  c(0, 4)

will also map c(0, 4)  h

(2,

8)

slide24
g

h

c(0, 4)

b(-2, 0)

a(4, 2)

The translation that maps a(4, 2)  c(0, 4)

will also map c(0, 4)  g

(-4,

6)

slide25
x1 = -2

y1 = 0

x2 = 0

y2 = 4

y2 – y1

Slope of bc =

x2 – x1

(4) – (0)

=

(0) – (-2)

4

=

2

b(-2, 0) c(0, 4)

= 2

slide26
b(-2, 0) c(0, 4) Slope = 2

x1 = 0 m = 2

y1 = 4

Equation of bc: y – y1 = m(x – x1)

y – 4 = 2(x – 0)

y – 4 = 2x

2x – y + 4 = 0 Equation of bc

coordinate geometry2

Coordinate Geometry

2000 Paper 2 Question 2

slide28
2x – y + 4 = 0 Equation of bc

h(2, 8)

When x = 2 and y = 8

2(2) – (8) + 4 =

0

 h(2, 8) is on the line bc

slide29
a(2, -3) b(-8, -6)

x1 = 2 x2 = -8

y1 = -3 y2 = -6

Mid-point of [ab] =

slide30
a(-2, -1), b(1, 0) c = (-5, 2)

x1 = -2 x2 = 1

y1 = -1 y2 = 0

slide31
a(-2, -1), b(1, 0) c = (-5, 2)

x1 = 1 x2 = -5

y1 = 0 y2 = 2

slide33
L: 3x – 4y + 20 = 0

To find the slope of L get y on it’s own

- 4y = - 3x - 20

Change Signs

Divide by 4

4y = 3x + 20

Slope of L = No. in front of x

y = ¾x + 5

Slope of L = ¾

slide34
Slope of K

x1 = 0

y1 = 5

y – (5) = (x – 0)

3y – 15 = x

Slope of L = ¾

K  L

Equation of K: y – y1 = m(x – x1)

3y – 15 = - 4x

Multiply by LCD = 3

4x + 3y – 15 = 0

K: 4x + 3y – 15 = 0

slide35
x = 15

x = - 20

4

3

15

- 20

, 0)

t = (

r = (

, 0)

4

3

K: 4x + 3y – 15 = 0

L: 3x – 4y + 20 = 0

L cuts x-axis when y = 0

K cuts x-axis when y = 0

4x + 3(0) – 15 = 0

3x – 4(0) + 20 = 0

4x = 15

3x = - 20

slide36
p(0, 5)

15

, 0)

r (

4

x 5

= ½ x

625

125

125

=

- 20

t (

, 0)

24

12

12

3

5

4

3

5

2

1

0

Area of ptr = ½ b.h

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