12 Coordinate Treatment of Simple Locus Problems

12 Coordinate Treatment of Simple Locus Problems

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12 Coordinate Treatment of Simple Locus Problems

Activity

Activity 12.1 (p. 37)

1. (a)

(b) (–2, 2), (0, 2), (3, 2), (4, 2) (or any other reasonable answers)

(c) 0x + y = 2 (i.e. y = 2)

2.  (a)

(b) (–3, –4), (–3, –1), (–3, 1), (–3, 4) (or any other reasonable answers)

(c) x + 0y = –3 (i.e. x = –3)

3. (a)

(b) (–2, –2), (0, 0), (1, 1), (4, 4) (or any other reasonable answers)

(c) y = x

4. (a)

(b) (–2, 4), (0, 2), (1, 1), (2, 0) (or any other reasonable answers)

(c) x + y = 2

Activity 12.2 (p. 46)

1. (a)

(b) ∵ = 2

∴ y = 2x – 1

2. (a)

(b) ∵ = m

∴ y – y1 = m(x – x1)

Activity 12.3 (p. 56)

1. (a) Ax + By + C = 0

By = –Ax – C

y =

(b) Slope =, y-intercept =

2.  By substituting (p, 0) into the equation Ax + By + C = 0,

we have:

A(p) + B(0) + C = 0

p =

∴ x-intercept =

3. Slope = 2, y-intercept = 1, x-intercept =

Activity 12.4 (p. 70)

1. = 3

∴ x2 + y2 = 32

∴ The equation of the circle is x2 + y2 = 9.

2. = 3

∴ (x – 1)2 + (y – 2)2 = 32

∴ The equation of the circle is (x – 1)2 + (y – 2)2 = 9.

3. = r

∴ (x – h)2 + (y – k)2 = r2

∴ The equation of the circle is (x – h)2 + (y – k)2 = r2.

Activity 12.5 (p. 82)

1. The straight line L1 cuts the y-axis at (0, c).

By the point-slope form,

y – c = m(x – 0)

∴ y = mx + c

2. Slope of L2 =

By the point-slope form,

y – y1 =(x – x1)

∴

3. The straight line L3 cuts the x-axis at (a, 0), and the y-axis at (0, b).

Slope of L3 =

=

By the point-slope form,

y – 0 =(x – a)

bx + ay = ab

∴

Follow-up Exercise

p. 43

1. The equation of L1 is y = 7.

The equation of L2 is y = –4.

2. The equation of L3 is x = 4.

The equation of L4 is x = –3.

3. The equation of L5 is y =x.

The equation of L6 is y = –x.

4. The equation of L7 is y =x.

The equation of L8 is y =x.

p. 49

1. The equation of the straight line L is

y – 3 =(x – 2)

∴ y =x + 2

2. The equation of the straight line L is

y – (–5) = –3[x – (–1)]

∴ y = –3x – 8

3. The equation of the straight line L is

y = 2x + 3

4. The equation of the straight line L is

y =x + 3

5. The equation of the straight line is

y – 3 =[x – (–2)]

∴ y =x

6. (a) The equation of the straight line is

y = 2x + 7

(b) The equation of the straight line is

y =x + (– 3)

∴ y =x – 3

p. 53

1. Let m be the slope of the straight line L.

m =

= 1

The equation of the straight line L is

y – 3 = 1(x – 1)

∴ y = x + 2

2.  Let m be the slope of the straight line L.

The equation of the straight line L is

y – 3 =(x – 1)

∴ y =

3.  Let m be the slope of the straight line L.

The equation of the straight line L is

y – 5 =(x – 0)

∴ y =x + 5

4.  Let m be the slope of the straight line L.

The equation of the straight line L is

y – (–2) =(x – 0)

∴ y =x – 2

5.  Let m be the slope of the straight line.

The equation of the straight line is

y – 8 =(x – 5)

∴ y =

6. Let m be the slope of the straight line.

The equation of the straight line is

y – (–5) =(x – 0)

∴ y =x – 5

p. 58

1. (a) 3y = –2x – 6

∴ 2x + 3y + 6 = 0

(b) 2(y – 3) = 4x

2y – 6 = 4x

4x – 2y + 6 = 0

∴ 2x – y + 3 = 0

(c) y – 1 = 3(x + 2)

y – 1 = 3x + 6

∴ 3x – y + 7 = 0

(d)

3y + 6 = –2x – 6

∴ 2x + 3y + 12 = 0

2. (a) From the equation 2x + y = 0, we have A = 2, B = 1 and C = 0.

∴ Slope =

=

=

x-intercept =

=

=

y-intercept =

=

=

Alternative Solution

Put y = 0 into 2x + y = 0, we have:

x = 0

∴ x-intercept =

If we make y the subject of the equation, we have:

y = –2x

∴ Slope =

y-intercept =

(b) From the equation 4x – 5y + 15 = 0, we have A = 4, B = –5 and C = 15.

∴ Slope =

=

=

x-intercept =

=

y-intercept =

=

=

Alternative Solution

Put y = 0 into 4x – 5y + 15 = 0, we have:

∴ x-intercept =

If we make y the subject of the equation, we have:

y =x + 3

∴ Slope =

y-intercept =

(c) From the equation 3x + 5y + 5 = 0, we have A = 3, B = 5 and C = 5.

∴ Slope =

=

x-intercept =

=

y-intercept =

=

=

Alternative Solution

Put y = 0 into 3x + 5y + 5 = 0, we have:

∴ x-intercept =

If we make y the subject of the equation, we have:

y =x – 1

∴ Slope =

y-intercept =

(d) From the equation 6x – 2y – 7 = 0, we have A = 6, B = –2 and C = –7.

∴ Slope =

=

=

x-intercept =

=

=

y-intercept =

=

Alternative Solution

Put y = 0 into 6x – 2y – 7 = 0, we have:

∴ x-intercept =

If we make y the subject of the equation, we have:

y = 3x –

∴ Slope =

y-intercept =

p. 62

1. From the equation of L: 3x – 4y + 10 = 0, we have:

Slope of L =

(a) ∵ L1 // L

∴ Slope of L1 = slope of L

=

∴ The equation of L1 is

y – 3 =(x – 1)

y =

(b) ∵ L2 ⊥ L

∴ Slope of L2 ´ slope of L = –1

Slope of L2 ´= –1

Slope of L2 =

∴ The equation of L2 is

y – 3 =(x – 0)

y =x + 3

2. From the equation of L: 2x + 5y – 7 = 0, we have:

Slope of L =

(a) ∵ L1 // L

∴ Slope of L1 = slope of L

=

∴ The equation of L1 is

y – 0 =[x – (–2)]

y =

(b) (i) ∵ L2 ⊥ L

∴ Slope of L2 ´ slope of L = –1

Slope of L2 ´ = –1

Slope of L2 =

∴ The equation of L2 is

y – 3 =(x – 0)

y =x + 3

(ii) Put y = 0 into y =x + 3, we have:

0 =x + 3

x =

∴ x-intercept =

p. 67

1. (a) L1: 3x + 5y – 1 = 0 ……(1)

L2: 2x – 5y + 1 = 0 ……(2)

(1) + (2),

(3x + 5y – 1) + (2x – 5y + 1) = 0

5x = 0

x = 0

By substituting x = 0 into (1), we have:

3(0) + 5y – 1 = 0

y =

∴ The coordinates of A =

(b) Slope of L =

∴ The equation of L is

y – 1 =(x – 1)

y =

2. (a) L1: 2x + y – 4 = 0 ……(1)

L2: 3x + y – 7 = 0 ……(2)

(2) – (1),

(3x + y – 7) – (2x + y – 4) = 0

x – 3 = 0

x = 3

By substituting x = 3 into (1), we have:

2(3) + y – 4 = 0

y = –2

∴ The coordinates of A =

(b) Slope of L =

∴ The equation of L is

y – 1 =(x – 1)

y =

3. (a) L1: 2x – 3y – 4 = 0 ……(1)

L2: 3x + y + 5 = 0 ……(2)

(1) + (2) ´ 3,

2x – 3y – 4 + 3(3x + y + 5) = 0

2x – 3y – 4 + 9x + 3y + 15 = 0

11x = –11

x = –1

By substituting x = –1 into (1), we have:

2(–1) – 3y – 4 = 0

y = –2

∴ The coordinates of A =

(b) Slope of L =

∴ The equation of L is

y – 1 =(x – 1)

y =

4. (a) L1: y = 3x + 1 ……(1)

L2: y = 5 – x ……(2)

By substituting (2) into (1), we have:

5 – x = 3x + 1

4x = 4

x = 1

By substituting x = 1 into (1), we have:

y = 3(1) + 1

= 4

∴ The coordinates of A =

(b) ∵ Notice that the x-coordinates of the points on L stay the same for different y-coordinates.

∴ L is parallel to the y-axis.

∴ The equation of L is x = 1.

p. 72

1. (a) x2 + y2 = 7

(b) (x – 4)2 + (y – 3)2 = 4

(c) x2 + (y + 3)2 = 16

(d) (x + 2)2 + (y – 5)2 = 8

(e) (x + 4)2 + (y + 5)2 =

2. (a) centre: (0, 0), radius =

(b) centre: (2, 3), radius = 6

(c) centre: (0, 1), radius = 5

(d) centre: (–1, –5), radius =

(e) centre: (0, 0), radius =

(f) centre: (–4, 0), radius =

3. For (x – 1)2 + (y – 1)2 = 9, centre: (1, 1), radius = 3

For (x + 1)2 + (y + 2)2 =, centre: (–1, –2), radius =

For (x – 2)2 + (y – 2)2 = 4, centre: (2, 2), radius = 2

∴ Matching is as follows:

p. 75

1. (x + 1)2 + (y – 1)2 = 16

x2 + 2x + 1 + y2 – 2y + 1 = 16

x2 + y2 + 2x – 2y – 14 = 0

2. (x – 7)2 + (y + 2)2 = 23

x2 – 14x + 49 + y2 + 4y + 4 = 23

x2 + y2 – 14x + 4y + 30 = 0

3. (x + 2)2 + y2 = 7

x2 + 4x + 4 + y2 = 7

x2 + y2 + 4x – 3 = 0

4.

5. (a) (x – 0)2 + (y – 0)2 = 62

x2 + y2 – 36 = 0

(b) (x – 0)2 + [y – (–4)]2 = 52

x2 + y2 + 8y – 9 = 0

(c) (x – 1)2 + (y – 2)2 = 32

x2 + y2 – 2x – 4y – 4 = 0

(d) [x – (–4)]2 + [y – (–5)]2 = 12

x2 + y2 + 8x + 10y + 40 = 0

(e) (x – 3)2 + (y – 0)2 = 22

x2 + y2 – 6x + 5 = 0

6. (a) Centre

Radius

(b) Centre

Radius

(c) Centre

Radius

(d) 2x2 + 2y2 – 4x – 7 = 0

x2 + y2 – 2x –= 0

Centre

Radius

(e) 4x2 + 4y2 + 12x – 20y + 8 = 0

x2 + y2 + 3x – 5y + 2 = 0

Centre

Radius

p. 79

1. (a) The coordinates of

(b) Radius