5ach12(coordinate Treatment Of Simple Locus Problems)

12 Coordinate Treatment of Simple Locus Problems

12

Coordinate Treatment of Simple Locus Problems

• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •

Activity

4.

(a)

Activity 12.1 (p. 37) 1.

(a)

(b) (–2, 4), (0, 2), (1, 1), (2, 0) (or any other reasonable answers) (c) x + y = 2 (b) (–2, 2), (0, 2), (3, 2), (4, 2) (or any other reasonable answers)

Activity 12.2 (p. 46)

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

1.

2.

(a)

(a)

y −5 x −3

(b) ∵

y −5 =2 x −3

∴ 2.

(a)

y − y1 x − x1

(b) ∵ (b) (–3, –4), (–3, –1), (–3, 1), (–3, 4) (or any other reasonable answers) (c) x + 0y = –3 (i.e. x = –3) 3.

(a)

y = 2x – 1

∴

y − y1 =m x − x1 y – y1 = m(x – x1)

Activity 12.3 (p. 56) 1.

(a) Ax + By + C = 0 By = –Ax – C y=−

(b) Slope = −

A C x− B B

A C , y-intercept = − B B

2. By substituting (p, 0) into the equation Ax + By + C = 0, we have: A(p) + B(0) + C = 0 p=−

(b) (–2, –2), (0, 0), (1, 1), (4, 4) (or any other reasonable answers) (c) y = x

∴

x-intercept = −

C A

C A 7

Certificate Mathematics in Action Full Solutions 5A

3.

Slope = 2, y-intercept = 1, x-intercept = −

1 2

2.

The equation of L3 is x = 4. The equation of L4 is x = –3.

3.

The equation of L5 is y =

Activity 12.4 (p. 70) 1.

( x −0)

∴ ∴ 2.

3.

+( y −0)

=3

2

2

+( y −2)

2

=3

∴ ∴

4.

(x – 1)2 + (y – 2)2 = 32 The equation of the circle is (x – 1)2 + (y – 2)2 = 9. ( x −h ) 2 +( y −k ) 2 = r

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

2 x. 3 5 The equation of L8 is y = − x. 4 The equation of L7 is y =

p. 49 1.

The equation of the straight line L is

2.

The straight line L1 cuts the y-axis at (0, c). By the point-slope form, y – c = m(x – 0) ∴ y = mx + c Slope of L2 =

y 2 − y1 x 2 − x1

∴ 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

By the point-slope form, y – y1 =

∴ 3.

y 2 − y1 (x – x1) x 2 − x1

y − y1 y − y1 = 2 x − x1 x 2 − x1

y=− 5.

b −0 0 −a b =− a

∴ 6.

By the point-slope form,

b (x – a) a

(a) The equation of the straight line is y = 2x + 7 (b) The equation of the straight line is

1 x + (– 3) 2 1 ∴ y= x–3 2 y=

bx + ay = ab

y x + =1 a b

Follow-up Exercise p. 43 1.

The equation of L1 is y = 7. The equation of L2 is y = –4.

3 [x – (–2)] 2 3 y=− x 2

y–3=−

Slope of L3 =

∴

2 x+3 3

The equation of the straight line is

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

y–0=−

1 (x – 2) 2 1 y= x+2 2

y–3=

Activity 12.5 (p. 82) 1.

1 x. 2

The equation of L6 is y = –x.

x2 + y2 = 32 The equation of the circle is x2 + y2 = 9. ( x −1)

∴ ∴

2

p. 53 1.

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

7 −3 5 −1

=1

8

12 Coordinate Treatment of Simple Locus Problems

The equation of the straight line L is y – 3 = 1(x – 1) ∴ y=x+2

∴ 6.

y=

1 27 x+ 4 4

Let m be the slope of the straight line.

−5 −0 0 −(−8) 5 =− 8

m= 2.

Let m be the slope of the straight line L.

4 −3 − 3 −1 1 =− 4

m=

The equation of the straight line is

5 (x – 0) 8 5 y=− x–5 8

y – (–5) = −

The equation of the straight line L is

1 (x – 1) 4 1 13 y=− x+ 4 4

∴

y–3=− ∴ 3.

p. 58 1.

∴

Let m be the slope of the straight line L.

0 −5 3 −0 5 =− 3

(b)

m=

4.

5 y – 5 = − (x – 0) 3 5 y=− x+5 3

0 − ( −2) 4 −0 1 = 2

m=

The equation of the straight line L is

1 (x – 0) 2 1 y= x–2 2

y – (–2) = ∴ 5.

∴

Let m be the slope of the straight line.

8 −6 5 −( −3) 1 = 4

m=

The equation of the straight line is y–8=

1 (x – 5) 4

∴ 2.

y – 1 = 3(x + 2) y – 1 = 3x + 6 3x – y + 7 = 0

y +2 2 =− x +3 3

(d)

Let m be the slope of the straight line L.

3y = –2x – 6 2x + 3y + 6 = 0

2(y – 3) = 4x 2y – 6 = 4x 4x – 2y + 6 = 0 ∴ 2x – y + 3 = 0

(c)

The equation of the straight line L is

∴

(a)

3y + 6 = –2x – 6 2x + 3y + 12 = 0

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

A B 2 =− 1

Slope = −

2 =−

C A 0 =− 2

x-intercept = −

=0

C B 0 =− 1

y-intercept = −

=0

9

Certificate Mathematics in Action Full Solutions 5A

=−

Alternative Solution Put y = 0 into 2x + y = 0, we have: x=0 ∴ x-intercept = 0 If we make y the subject of the equation, we have: y = –2x 2 ∴ Slope = −

x-intercept = −

C B 5 =− 5

A B  4  = −   −5 

Slope = −

C B  15  = −   −5 

y-intercept = −

=3 Alternative Solution Put y = 0 into 4x – 5y + 15 = 0, we have:

4 x − 5(0) + 15 = 0 4 x = −15 15 x =− 4

15 x-intercept = − 4

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

4 x+3 5

Slope =

4 5

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

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

A Slope = − B

3 x + 5(0) + 5 = 0 3 x = −5 5 x =− 3

C A

15 =− 4

∴

1 =−

4 5

x-intercept = −

5 3

y-intercept = −

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

=

C A

=−

y-intercept = 0

∴

3 5

∴

x-intercept = −

5 3

If we make y the subject of the equation, we have: y=− ∴

3 x–1 5

Slope = −

3 5

1 y-intercept = −

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

B

A B  6  = −   −2 

Slope = −

=3

C A  −7  = −   6 

x-intercept = −

=

7 6

y-intercept = −

C B

 −7  = −  −2  10

12 Coordinate Treatment of Simple Locus Problems

=−

7 2

Slope of L = − (a) ∵ ∴

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

6 x − 2( 0) − 7 = 0 6x = 7 7 x= 6

∴

x-intercept =

L1 // L Slope of L1 = slope of L =−

∴

∴

2 [x – (–2)] 5 2 4 y=− x− 5 5

y–0=−

7 6

7 2

(b) (i) ∵ ∴

Slope of L2 ×

∴

p. 62

(a) ∵ ∴

5 (x – 0) 2 5 y= x+3 2 5 (ii) Put y = 0 into y = x + 3, we have: 2 5 0= x+3 2 6 x=− 5

3 3 = ( −4) 4

L1 // L Slope of L1 = slope of L =

∴

3 4

The equation of L1 is

3 (x – 1) 4 3 9 y= x+ 4 4

y–3=

(b) ∵ ∴

L2 ⊥ L Slope of L2 × slope of L = –1

3 Slope of L2 × = –1 4 4 Slope of L2 = − 3 ∴

The equation of L2 is

4 (x – 0) 3 4 y=− x+3 3

y–3=−

2.

The equation of L2 is y–3=

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

L2 ⊥ L Slope of L2 × slope of L = –1

 2  −  = –1  5 5 Slope of L2 = 2

Slope = 3

7 y-intercept = − 2

1.

2 5

The equation of L1 is

If we make y the subject of the equation, we have: y = 3x –

2 5

∴

x-intercept = −

6 5

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= ∴

1 5

The coordinates of A = (0,

1 ) 5

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

11

Certificate Mathematics in Action Full Solutions 5A

4x = 4 x=1 By substituting x = 1 into (1), we have: y = 3(1) + 1 =4 ∴ The coordinates of A = (1, 4)

1 (b) Slope of L = 5 =4 1−0 5 1−

∴

The equation of L is

4 (x – 1) 5 4 1 x+ y= 5 5

y–1=

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 = (3, −2)

(b) ∵ ∴ ∴

p. 72 1.

The equation of L is

3 y – 1 = − (x – 1) 2 3 5 y=− x+ 2 2 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 1, − 2) ∴ The coordinates of A = ( − (b) Slope of L = ∴

(c) x2 + (y + 3)2 = 16 (d) (x + 2)2 + (y – 5)2 = 8 (e) (x + 4)2 + (y + 5)2 = 2.

3.

9 4

(a)

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

(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 2

10

4 3 3

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

9 3 , centre: (–1, –2), radius = 4 2

For (x – 2)2 + (y – 2)2 = 4, centre: (2, 2), radius = 2 ∴ Matching is as follows:

1 −( −2) 3 = 1 −(−1) 2

The equation of L is

3 (x – 1) 2 3 1 y= x− 2 2

y–1=

4.

(a) x2 + y2 = 7 (b) (x – 4)2 + (y – 3)2 = 4

1 −(−2) 3 =− (b) Slope of L = 1 −3 2 ∴

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.

(a) L1: y = 3x + 1 ……(1) L2: y = 5 – x ……(2) By substituting (2) into (1), we have: 5 – x = 3x + 1

p. 75 1.

(x + 1)2 + (y – 1)2 = 16 x + 2x + 1 + y2 – 2y + 1 = 16 x2 + y2 + 2x – 2y – 14 = 0 2

2.

(x – 7)2 + (y + 2)2 = 23 x2 – 14x + 49 + y2 + 4y + 4 = 23 x2 + y2 – 14x + 4y + 30 = 0

12

12 Coordinate Treatment of Simple Locus Problems

3.

4.

(x + 2)2 + y2 = 7 x2 + 4x + 4 + y2 = 7 x2 + y2 + 4x – 3 = 0

Centre

( x + 2 ) 2 +( y + 3 ) 2 =17

2

2

x 2 + y 2 +2 2 x +2 3 y −12 =0

5.

=

(a) (x – 0)2 + (y – 0)2 = 62 x2 + y2 – 36 = 0

(c)

2

( −5)   3 =− ,−  2 2   Centre  3 5 =− ,   2 2

2

(x – 1) + (y – 2) = 3 x2 + y2 – 2x – 4y – 4 = 0

(d) [x – (–4)]2 + [y – (–5)]2 = 12 x2 + y2 + 8x + 10y + 40 = 0

2

=

6  ( −4) =− ,−  (a) Centre 2 2  =( 2,−3) 2

2

1.

0  8 =− ,−  2  2 =( −4,0) 2

∴

2

= 19

2.

( −12 )   0 =− ,−  2   2 =(0,6) 2

2

0   −12  Radius =  2  +  2  − 20     =4 (d) 2x2 + 2y2 – 4x – 7 = 0 x2 + y2 – 2x –

7 =0 2

(a) The coordinates of

1 +5 −4 +( −6)  C = ,  2  2  =(3,−5)

2 2 (b) Radius = (1 −3) +[ −4 −( −5)] = 5

8  0   +  −( −3) Radius =  2   2 

(c) Centre

26 2

p. 79

 −4  6  Radius =  2  +  2  − ( −12 )     =5 (b) Centre

2

3  −5  =   +  −2 2    2  Radius

(e) (x – 3)2 + (y – 0)2 = 22 x2 + y2 – 6x + 5 = 0

6.

3 2 2

(e) 4x2 + 4y2 + 12x – 20y + 8 = 0 x2 + y2 + 3x – 5y + 2 = 0

(b) (x – 0)2 + [y – (–4)]2 = 52 x2 + y2 + 8y – 9 = 0 2

2

 −2  0  7 =   +   − −  2 2      2 Radius

x +2 2 x +2 + y +2 3 y +3 =17 2

0  ( −2) =− ,−  2 2  = (1,0)

The equation of the circle is (x – 3)2 + [y – (–5)]2 = ( 5 ) 2 2 x + y2 – 6x + 10y + 29 = 0

(a) Let the equation of the circle be: x2 + y2 + Dx + Ey + F = 0 ……(1) By substituting (1, 2) into (1), we have: 12 + 22 + D(1) + E(2) + F = 0 i.e. D + 2E + F = –5 ……(2) By substituting (–2, 3) into (1), we have: (–2)2 + 32 + D(–2) + E(3) + F = 0 i.e. –2D + 3E + F = –13 ……(3) By substituting (0, –4) into (1), we have: 02 + (–4)2 + D(0) + E(–4) + F = 0 i.e. –4E + F = –16 ……(4) (2) – (3), 3D – E = 8 ……(5) (3) – (4), –2D + 7E = 3 ……(6) (5) × 7 + (6), 19D = 59 D=

59 19 13

Certificate Mathematics in Action Full Solutions 5A

By substituting D =

y − (−3) −4 − ( −3) = x −2 −1 − 2 y +3 −1 = x − 2 −3 3 y +9 = x − 2 x − 3 y −11 = 0

59 into (5), we have: 19

 59  –E=8  19  25 E= 19 25 By substituting E = into (4), we have: 19  25  –4   + F = –16  19  204 F=− 19 3

∴

(b) The equation of the straight line is

y −( −5) −6 −( −5) = x −( −4) −3 −( −4) y +5 −1 = x +4 1 y +5 = −x −4

The equation of the circle is

59 25 204 x +y + x+ y− = 0. 19 19 19 2

x + y +9 = 0

2

(b) Let the equation of the circle be: x2 + y2 + Dx + Ey + F = 0 ……(1) By substituting (1, 0) into (1), we have: 12 + 02 + D(1) + E(0) + F = 0 i.e. D + F = –1 ……(2) By substituting (6, 0) into (1), we have: 62 + 02 + D(6) + E(0) + F = 0 i.e. 6D + F = –36 ……(3) By substituting (0, 3) into (1), we have: 02 + 32 + D(0) + E(3) + F = 0 i.e. 3E + F = –9 ……(4) (2) – (3), – 5D = 35 D = –7 (3) – (4), 6D – 3E = –27 2D – E = –9 ……(5) By substituting D = –7 into (5), we have: 2(–7) – E = –9 E = –5 By substituting D = –7 into (2), we have: –7 + F = –1 F=6 ∴ The equation of the circle is x2 + y2 – 7x – 5y + 6 = 0.

3.

(a) The equation of the straight line is

x y + =1 3 −1 x −3 y − 3 = 0 (b) The equation of the straight line is

x y + =1 1 −2     2 x − 4y + 2 = 0

Exercise Exercise 12A (p. 34) Level 1 1.

(a)

AB = [3 −( −1)] 2 +(0 −3) 2 = 4 2 +( −3) 2 =5

0 −3 3 −( −1) 3 =− 4

Slope of AB =

p. 85 1.

(a) The equation of the straight line is y = 2x – 1 2x – y – 1 = 0 (b) The equation of the straight line is

2 y=− x+4 3

Let θ be the inclination of AB. ∵ Slope of AB = tan θ ∴

2x + 3y – 12 = 0 2.

(a) The equation of the straight line is

∴

−

3 = tan θ 4 θ =180 ° −36 .87 ° =143 .1° (cor. to 1 d.p.)

The inclination of AB is 143.1°.

14

12 Coordinate Treatment of Simple Locus Problems

∴

PQ = ( −2 −4) 2 +( −1 −2) 2

(b)

Slope of PQ = slope of QR

−1 −4 −8 −( −1) = 3 −2 x −3 5 x −15 = 7

= ( −6) 2 +( −3) 2 =3 5

−1 − 2 −2 −4 1 = 2

Slope of PQ =

x= 5.

Let θ be the inclination of PQ. ∵ Slope of PQ = tan θ ∴

(a) ∵ ∴

AB // CD Slope of AB = slope of CD

7 −2 y − ( −1) = 4 −0 7 −2 25 = 4 y + 4 21 y= 4

1 =tan θ 2 θ =26.6 ° (cor. to 1 d.p.) ∴

The inclination of PQ is 26.6°.

(b) ∵ ∴

MN = [ −3 −( −1)] 2 +[ 2 −( −4)] 2

(c)

= ( −2) 2 +6 2 =2 10

2 −( −4) −3 −( −1) = −3

22 5

AC ⊥ BD Slope of AC × slope of BD = –1 −1 −2 y −7 × = −1 2 −0 7 −4 y −7 = 2

y =9

Slope of MN =

6.

Let θ be the inclination of MN. ∵ Slope of MN = tan θ ∴

(a) Let (0, y) be the coordinates of C. ∵ A, B and C are collinear. ∴ Slope of AC = slope of AB

y −6 3 −6 = 0 −( −2) −3 −( −2) y −6 =3 2 y =12

−3 =tan θ

θ =180 °−71.57 ° =108 .4° (cor. to 1 d.p.) ∴ The inclination of MN is 108.4°. 2.

∴

Perimeter of ABC = AB + BC + CA =

( 26 −1) +(1 −1) 2

2

The coordinates of C =(0, 12 )

(b) Let θ be the inclination of AC. ∵ Slope of AC = 3 (from (a)) ∴ tan θ = 3 θ = 71.6° (cor. to the nearest 0.1°) ∴ The inclination of AC is 71.6°.

+

(10 −26 ) 2 +(13 −1) 2 + (1 −10 ) 2 +(1 −13 ) 2

= 25 + 20 + 15 = 60 3.

Let (x, 0) be the coordinates of B. ∵ AB = 5 ∴

( x −5) 2 +(0 −4) 2 = 5 2

4.

∴

(x – 5) = 9 x=8 or x = 2 The coordinates of B = (8, 0) or ( 2, 0)

∵

P, Q and R are collinear.

7.

(a) Let (x, y) be the coordinates of M. ∵ M is the mid-point of AB. ∴

∴

x=

0 +4 and 2

y=

1 +7 2

x=2 and y=4 The coordinates of M = ( 2, 4)

(b) Let (0, c) be the coordinates of C.

15

Certificate Mathematics in Action Full Solutions 5A

∵ ∴

4( −5) +3( 2) 3 +4 = −2 4(6) +3( −2) y= 3 +4 18 = 7

CM ⊥ AB Slope of CM × slope of AB = –1

x=

c − 4 7 −1 × = −1 0 −2 4 −0 4 c −4 = 3 16 c= 3 ∴ 8.

∴

2 +x = 5 and 2

3(−13 ) + 7(7) 7 +13 =1 3(0) + 7(10 ) y= 7 +3 =7 x=

4+y =2 2

x=8 and y=0 ( 8 , 0) The coordinates of B =

(b) (i) Slope of OA = ∵ ∴

4 −0 2 −0

∴

=2 CM // OA Slope of CM = slope of OA =2

∴ 9.

(1, 7 )

s ( −1) + r ( −6) r +s − 4r − 4 s = − s − 6r 2r = 3s −4 =

2− y =2 5 −0 y = −8 The coordinates of C =

The coordinates of P =

10. (a) Let AP : PB = r : s. By the section formula for internal division, we have:

(ii) Let (0, y) be the coordinates of C. ∵ Slope of CM = 2 ∴

18 ) 7

(c) Let (x, y) be the coordinates of P. By the section formula for internal division, we have:

(a) Let (x, y) be the coordinates of B. ∵ M is the mid-point of AB. ∴

The coordinates of P = (−2,

∴

16 ) The coordinates of C = (0, 3

r 3 = s 2 AP : PB = 3 : 2

∴ (0, − 8)

(b) By the section formula for internal division, we have:

2( −2) +3(7) 3 +2 17 = 5

y=

(a) Let (x, y) be the coordinates of P. By the section formula for internal division, we have:

1(4) + 2(7) 2 +1 =6

x=

1(6) + 2(9) y= 2 +1 =8 ∴ The coordinates of P = (6, 8) (b) Let (x, y) be the coordinates of P. By the section formula for internal division, we have:

11. PA = ∵ ∴

( 2 −a ) 2 +(3 −b) 2

PA is an integer. Point P is (2, 4), (2, 5), (3, 3) or (4, 3) (or any other reasonable answers).

12. Let (x1, y1), (x2, y2), (x3, y3) and (x4, y4) be the coordinates of A, B, C and D respectively. ∵ ABCD is a parallelogram. ∴ K is the mid-point of AC and BD. ∴

1=

x1 + x3 y + y3 ,2= 1 2 2

and

16

12 Coordinate Treatment of Simple Locus Problems

1=

x2 + x 4 y + y4 ,2= 2 2 2

x1 + x3 = 2, ∴

∴

y1 + y3 = 4 and x2 + x4 = 2, y2 + y4 = 4

The possible coordinates of A, B, C and D are (3, 5), (0, 4), (–1, –1) and (2, 0) or (4, 4), (–1, 3), (–2, 0) and (3, 1) respectively (or any other reasonable answers).

c −5 = −8 c = −3

(ii) Area of ABC

Level 2

1 × AB × BC 2 1 = × [3 − ( −1)] 2 + (5 − 4) 2 × (5 − 3) 2 + ( −3 − 5 2 1 = × 17 × 2 17 2 = 17 =

13. (a) Let (0, y) be the coordinates of C. ∵ CB // OA ∴ Slope of CB = slope of OA

10 − y 6 − 0 = 5 −0 8 −0 40 − 4 y = 15 4 y = 25 25 y= 4

∴

The coordinates of C = (0,

25 ) 4

(b) Let (x, 0) be the coordinates of D. ∵ BD ⊥ AC ∴ Slope of BD × slope of AC = –1

5 − 0 4 − (−3) × = −1 3 − x −1 − 5 5 6 = 3− x 7 35 = 18 − 6 x

(b) Slope of AB × slope of OA

10 − 6 6 × 5 −8 8 4 3 = × −3 4 = −1 =

∴ (c)

x=−

AB ⊥ OA

∴

OA = (8 −0) 2 +(6 −0) 2

AB = (5 −8) 2 + (10 − 6) 2

y 2 +8 y +16 = 64 + y 2 8 y = 48 y =6

2

25   CB = (5 − 0) 2 + 10 −  4   25 = 4

1 × (CB + OA ) × AB 2 1  25  = × +10  ×5 2  4  = 40 .625 14. (a) (i) ∵

17 , 0) The coordinates of D = (− 6

y − ( −4) = (8 − 0) 2 + (0 − y ) 2

=5

=

17 6

15. (a) Let (0, y) be the coordinates of A. ∵ AB = AC

=10

Area of trapezium OABC

Slope of AB × slope of BC = –1 5 −4 c −5 × = −1 3 −( −1) 5 −3

∴

The coordinates of A = (0, 6)

 0 +8 −4 + 0  ,  2   2 = ( 4,−2)

(b) The coordinates of M =  Slope of AM × slope of BC

AB ⊥ BC

17

Certificate Mathematics in Action Full Solutions 5A

−2 − 6 0 − ( −4) × 4 −0 8 −0 −8 4 = × 4 8 = −1

2

∴

2

 1  5   3 9  PQ =  −  −  +  −   2  2   2 2 

=

= 32 + (−3) 2 =3 2

AM ⊥ BC

2

 5  1   7 13  SR =  −  −  +  −  2  2  2  2 

(c) Area of ABC

2

= 32 + ( −3) 2

1 =3 2 × BC × AM 2 ∴ PS = QR and PQ = SR 1 = × (8 − 0) 2 +[0 − ( −4)] 2 × ( 4 − 0) 2 + ( −2 − 6) 2 (ii) Slope of PS × slope of PQ 2 13 9 3 9 1 − − = × 80 × 80 2 2 × 2 2 2 = 1  5 1  5 = 40 − −−  −−  2  2 2  2   −2 +(−3 ) 6 +3  P = ,  2 −3 2 2   = × 16. (a) The coordinates of 2 3  5 9 = − ,  = − 1  2 2 =

∴ PS ⊥ PQ Slope of QR × slope of SR

 −3 +4 3 + 0  Q = ,  2   2 The coordinates of 1 3  = ,  2 2 

4 +1 0 + 7  The coordinates of R =  ,  

2   2 5 7  = ,  2 2  1 +(−2 ) 7 + 6  S = ,  2 2   The coordinates of  1 13  = − ,   2 2  2

(b) (i)

 1  5  13 9  PS = − −  −  +  −  2 2    2 2  = 22 + 22 = 2 2 2

5 1 7 3  QR =  −  +  −   2 2 2 2 = 22 + 22 = 2 2

2

7 3 7 13 − − = 2 2× 2 2 5 1 5  1 − −−  2 2 2  2 2 −3 = × 2 3 = −1

∴

QR ⊥ SR

17. (a) (i) Let (0, y) be the coordinates of D. ∵ D lies on AB. ∴ Slope of AD = slope of AB

y −6 −2 − 6 = 0 − 2 −3 − 2 16 y −6 = − 5 14 y= 5

2

∴

The coordinates of D = (0,

14 ) 5

(ii) Let AD : DB = r : s. By the section formula for internal division, we have:

18

12 Coordinate Treatment of Simple Locus Problems

2 − 7 3 − ( −2) × = −1 5 − a 7 − ( −3) 5 = 10 − 2a 5 a= 2

s ( 2) + r ( −3) r +s 2 s = 3r 0=

r 2 = s 3 ∴

AD : DB = 2 : 3

(b) (i) Slope of AC = ∵ ∴

2

5  PH = 5 −  + ( 2 − 7) 2 2 

2 −6 4 =− 7 −2 5

=

DE // BC Slope of DE = slope of BC

2 −( −2) 7 −( −3) 2 = 5

(c) Area of PQR

=

(ii) Let (x, y) be the coordinates of E. ∵ Slope of AE = slope of AC ∴

∴

2 5

……(2)

22 . 5

22 ) ∴ The coordinates of E = ( 4, 5 18. (a) Let (x, y) be the coordinates of H. By the section formula for internal division, we have:

1( −3) + 4(7) x= 4 +1 =5 1( −2) + 4(3) y= 4 +1 =2 (b) ∵ ∴

The coordinates of H =

1 5 5 × [7 −( −3)] 2 +[3 −( −2)] 2 × 2 2

Level 1

By solving (1) and (2), we have x = 4, y =

∴

=

Exercise 12B (p. 44)

14 y− 5 =2 x −0 5 2x – 5y + 14 = 0

1 ×QR ×PH 2

1 5 5 ×5 5 × 2 2 125 = 4

y −6 4 =− x −2 5

Slope of DE =

=

=

5y – 30 = –4x + 8 4x + 5y – 38 = 0 ……(1) ∵

5 5 2

1.

∵ ∴

L is parallel to the x-axis. The equation of L is y = 3.

2.

∵ ∴

L is parallel to the x-axis. The equation of L is y = –5.

3.

∵ ∴

L is parallel to the y-axis. The equation of L is x = 6.

4.

∵ ∴

L is parallel to the y-axis. The equation of L is x = –2.

5.

The equation of L is

6 x 2 i.e. y = 3x y=

6.

The equation of L is

2 x −4 1 y =− x 2 y=

(5, 2)

i.e.

PH ⊥ QR Slope of PH × slope of QR = –1 7.

∵ ∴

Slope = 2 The equation of L is y = 2x.

19

Certificate Mathematics in Action Full Solutions 5A

8.

9.

1 3

∵

Slope = −

∴

The equation of L is y = −

By substituting (a, 6) into the equation y = −

1 x. 3

have: 6=−

3 a 4

8 a= −

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

∴

(b) The equation of the straight line is y = 0.

By substituting (6, b) into the equation y = −

(c) The equation of the straight line is y = –3.

have:

(d) The equation of the straight line is y = 7. 10. (a) The equation of the straight line is x = 4.

∴

b=−

3 (6) 4

b=−

9 2

(b) The equation of the straight line is x = –2. (c)

The equation of the straight line is x = –5.

3 x , we 4

3 x , we 4

14. (a)

(d) The equation of the straight line is x = 0. 11. (a) The equation of the straight line is y=

8 x 2

i.e. y = 4x (b) The equation of the straight line is y= i.e. y =

−5 x −4

5 x 4

(b) (i) By joining the points in (a), we can see that L passes through the origin. ∴

(c) The equation of the straight line is y=

4.5 x −1.5

∴

i.e. y = –3x (d) The equation of the straight line is y= i.e. y = −

−9 x 12

3 x 4

Level 2 12. By substituting (1, a) into the equation y = 2, we have: a =2 By substituting (3, b) into the equation y = 2, we have: b =2 By substituting (–4, c) into the equation y = 2, we have: c =2 13. ∵

L passes through A and B.

y − coordinate s is a constant. x − coordinate s y 3 = x 2 y : x = 3: 2

(ii)

y 3 = x 2 3 y= x 2 3 ∴ The equation of L is y = x . 2 From (b) (i), we have

15. (a) Let (0, y), (h, k) and (x, 0) be the coordinates of A, B and C respectively. By substituting the points (0, y) and (h, k) into the equation y = 3, we have y = k = 3. ∵ OABC is a square. ∴ OC = k = 3 ∵ C is on the negative x-axis. ∴ x = –3 CB = OC = 3

20

12 Coordinate Treatment of Simple Locus Problems

∵ ∴ ∴

B is in Quad. II. h = –3 The coordinates of A, B and C are (0, 3), (–3, 3) and (–3, 0) respectively.

(b) The equation of OB is

3 x, −3 i.e. y = −x y=

4.

The equation of the straight line L is y = 2x + 2

5.

The equation of the straight line L is y = –5x + 7

6.

The equation of the straight line L is

∴

16. (a) The equation of L1 is

−2 x, −3 i.e. 2 y= x 3 y=

The equation of L2 is i.e.

−3 y= x, 1 y = −3 x

7.

The equation of the straight line is y – 2 = 4(x – 3) ∴ y = 4x – 10

8.

The equation of the straight line is y – 3 = –2[x – (–2)] ∴ y = –2x – 1

9.

The equation of the straight line is

2 3 θ1 = 33 .7° =

∴

10. The equation of the straight line is



= 108 .4° θ2 – θ1 = 108.4° – 33.7° = 75° (cor. to the nearest degree) ∴ The acute angle between L1 and L2 is 75°.



∴

y=3

11. The equation of the straight line is y = 3x + 4 12. The equation of the straight line is y=−

Exercise 12C (p. 50)

2.

3.

The equation of the straight line L is y – 4 = 3(x – 0) ∴ y = 3x + 4 The equation of the straight line L is y – 3 = –1[x – (–2)] ∴ y = –x + 1 The equation of the straight line L is

1 [x – (–2)] 2 1 y= x+5 2

y–4= ∴

2 x+1 5

13. The equation of the straight line is

Level 1 1.

 1    3 

y – 3 = 0x −  −

tan θ2 = slope of L2 = −3 θ2 = 180 ° − 71 .6°

4 [x – (–2)] 5 4 17 y=− x+ 5 5

y–5=−

(b) Let θ1 and θ2 be the inclinations of L1 and L2 respectively.

tan θ1 = slope of L1

1 x + (–3) 2 1 y=− x–3 2 y=−

y=

4 4 x+ 7 7

14. The equation of the straight line is y = 0x + (–4) ∴ y = –4 15. Let the slope of the straight line be m. The equation of the straight line L is y – 3 = m[x – (–2)] y = mx + (2m + 3) Let m = –1. y = (–1)x + [2(–1) + 3] = –x + 1 Let m = –2. y = (–2)x + [2(–2) + 3] = –2x – 1 ∴ The possible equation of a straight line L is

21

Certificate Mathematics in Action Full Solutions 5A

y = –x + 1 or y = –2x – 1 (or any other reasonable answers). 16. Let the slope and y-intercept of the straight line L be m and c respectively. The equation of the straight line L is y = mx + c Let m = 1 and c = –1. y = (1)x + (–1) = x – 1 Let m = 2 and c = –5. y = (2)x + (–5) = 2x – 5 ∴ The possible equation of a straight line L is y = x – 1 or y = 2x – 5 (or any other reasonable answers).

Level 2 17. (a) The equation of the straight line L is y – 2 = 1[x – (–2)] ∴ y=x+4 (b) By substituting B(2, –6) into the equation y = x + 4, we have: L.H.S. = –6 R.H.S. = 2 + 4 = 6 ≠ –6 ∴ B does not lie on L. By substituting C(3, 7) into the equation y = x + 4, we have: L.H.S. = 7 R.H.S. = 3 + 4 = 7 = L.H.S. ∴ C lies on L.

(c) ∵

Slope of L3 =

= ∴

= –2 The equation of L3 is y = –2x + (–4) y = –2x – 4

20. (a) (i) The equation of the straight line L1 is y – 10 = 2(x – 3) ∴ y = 2x + 4 (ii) From the equation of L1: y = 2x + 4, y-intercept c = 4

3 × slope of L1 4 3 = × 2 4 3 = 2

(b) (i) Slope of L2 =

y-intercept = c + 2 = 4 + 2 = 6 ∴

(b) By substituting (3, –1) into the equation y = mx + 5, we have: –1 = m(3) + 5 2 m= −

y-intercept = 3

(b) ∵

Slope of L2 = 4 × slope of L1

 1 = 4 × −   2 ∴

= –2 The equation of L2 is y – (–1) = –2[x – (–2)] y = –2x – 5

3 x + 6. 2

3 x + 6 , we have: 2 3 a= (2a) + 6 2

y=

y–4=−

(ii) From the equation of L1: y = −

The equation of L2 is y =

(ii) By substituting (2a, a) into the equation

19. (a) (i) The equation of the straight line L1 is

∴

1

1 −   2

18. (a) The equation of the straight line L is y = mx + 5.

1 [x – (–2)] 2 1 y=− x+3 2

1 slope of L1

3 a= −

Exercise 12D (p. 53) Level 1

1 x +3 , 2

1.

Let m be the slope of the straight line L.

1 −3 2 −1 = −2

m=

The equation of the straight line L is y – 1 = –2(x – 2) ∴ y = –2x + 5

2.

Let m be the slope of the straight line L.

22

12 Coordinate Treatment of Simple Locus Problems

1 − ( −1) 3 − ( −2) 2 = 5

0 − ( −3) −5 − 0 3 =− 5

m=

m=

The equation of the straight line L is

The equation of the straight line L is

∴ 3.

2 y–1= (x – 3) 5 2 1 x− y= 5 5 Let m be the slope of the straight line L.

−7 −(−8) 6 −(−7) 1 = 13

3 (x – 0) 5 3 y=− x–3 5

y – (–3) = − ∴ 7.

Let m be the slope of the straight line.

0 − ( −2) 3 − ( −1) 1 = 2

m=

m=

The equation of the straight line L is

The equation of the straight line is

1 (x – 6) 13 1 97 x− y= 13 13

∴ 4.

Let m be the slope of the straight line L.

0 −3 m= 6 −0 1 =− 2 The equation of the straight line L is

1 (x – 0) 2 1 y=− x+3 2

y–3=− ∴ 5.

Let m be the slope of the straight line L.

2 −0 m= 0 −(−5) 2 = 5 The equation of the straight line L is

2 (x – 0) 5 2 y= x+2 5

∴ 8.

6.

Let m be the slope of the straight line.

−1 −(−3) 2 −1 =2

m=

The equation of the straight line is y – (–1) = 2(x – 2) ∴ y = 2x – 5 9.

Let m be the slope of the straight line.

−6 −( −5) −3 −( −4) = −1

m=

The equation of the straight line is y – (–6) = –1[x – (–3)] ∴ y = –x – 9 10. Let m be the slope of the straight line.

1 −( −7) 1 −( −5) 4 = 3

m=

y–2= ∴

1 (x – 3) 2 1 3 y= x− 2 2

y–0=

y – (–7) =

The equation of the straight line is

4 (x – 1) 3 4 1 y= x− 3 3

y–1=

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

23

Certificate Mathematics in Action Full Solutions 5A

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

15. (a)

m=

The equation of the straight line is

∴

The equation of the straight line L is

1 y – 1 = − (x – 0) 3 1 y=− x+1 3

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

5 −0 2 m= 0 −5 1 =− 2 The equation of the straight line is y– ∴

1 5 = − (x – 0) 2 2 1 5 y=− x+ 2 2

Let m be the slope of the straight line L.

6 − ( −2) −5 −5 4 =− 5

1 −0 m= 0 −3 1 =− 3

∴

(b) From the equation of L: y = −

5 −0 0 − ( −4) 5 = 4

16. (a)

Let m be the slope of the straight line L.

−2 − 0 0 −a 2 = a

m=

The equation of the straight line L is

5 ( x − 0) 4 5 y = x +5 4

∴

∴

(b) ∵ ∴

17. (a) The equation of the straight line L1 is

4 [x – (–2)] 3 4 1 y= x− 3 3

y – (–3) =

m=

The equation of the straight line L2 is

The equation of the straight line is

4 (x – 0) 3 4 y=− x–4 3

y – (–4) = − ∴

1 3 2 1 = a 3

Slope of L =

a=6

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

−4 − 0 0 − ( −3) 4 =− 3

2 x + (–2) a 2 y= x–2 a y=

The equation of the straight line is

y −5 =

4 x + 2, 5

y-intercept = 2

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

m=

4 (x – 5) 5 4 y=− x+2 5

y – (–2) = −

4 (x – 4) 3 4 10 x− y= 3 3

y–2= ∴

(b) (i) Let m be the slope of the straight line.

Level 2 24

12 Coordinate Treatment of Simple Locus Problems

2 −( −3) 4 −( −2) 5 = 6

m=

∴

The equation of the required straight line is

∴

C A x-intercept 24 =− 8 = −3 =−

5 y–2= (x – 4) 6 5 4 y= x− 6 3

C B y-intercept 24 =− 3 = −8 =−

(ii) From the equation of the required straight line: y=

5 4 x− , 6 3

y-intercept = −

4 3

Alternative Solution

18. Let the x-intercept be 2a, then the y-intercept is a. Let m be the slope of the straight line.

Put y = 0 into 8x + 3y + 24 = 0, we have:

8 x + 3(0) + 24 = 0 8 x = −24

a −0 0 − 2a 1 =− 2

m=

x = −3

3 x-intercept = −

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

The equation of the straight line is

∴

y=−

1 y – 2 = − (x – 1) 2 1 5 y=− x+ 2 2

∴

6.

Level 1 ∴

4y = –3x + 2 3x + 4y – 2 = 0

∴

4(y – 1) = 2x +1 4y – 4 = 2x +1 2x – 4y + 5 = 0

∴

y – 3 = –2(x +1) y – 3 = –2x – 2 2x + y – 1 = 0

2.

3.

∴ 5.

A B Slope 7 =− 2 =−

x-intercept = − C

A  −14  = −C  = − 7  B y-intercept = 2 −14  = −   2  =7

2y – 8 = –x + 2 x + 2y – 10 = 0

From the equation 8x + 3y + 24 = 0, we have A = 8, 3 and C = 24.

8 Slope = − 3

From the equation 7x + 2y – 14 = 0, we have A = 7, B = 2 and C = –14. ∴

y −4 1 =− x −2 2

4.

8 x–8 3

8 y-intercept = −

Exercise 12E (p. 59) 1.

A B Slope 8 =− 3 =−

B=

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

25

Certificate Mathematics in Action Full Solutions 5A

y=− ∴

7 x+7 2

∴

7 Slope = − 2

y-intercept = 7 Put y = 0 into 7x + 2y – 14 = 0, we have:

C A −2  x-intercept = −    3  2 = 3 =−

7 x + 2(0) −14 = 0

7 x =14 x =2 ∴ 7.

x-intercept = 2

From the equation 2x – 3y + 9 = 0, we have A = 2, B = –3 and C = 9.

C B −2  y-intercept = −    7  2 = 7 =−

A B  Slope = − 2    −3  2 = 3 =−

∴

A B Slope 3 =− 7 =−

Alternative Solution

C =− A x-intercept 9 =− 2

If we make y the subject of the equation, we have: y=− ∴

C B y-intercept  9  = −   −3  =3

3 2 x+ 7 7

Slope = −

=−

y-intercept =

3 x + 7 ( 0) − 2 = 0 3x = 2 2 x= 3

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

∴

2 x+3 3

Slope =

2 3

y-intercept = 3 Put y = 0 into 2x – 3y + 9 = 0, we have:

2 x − 3(0) + 9 = 0 2 x = −9 9 x =− 2

∴ 8.

x-intercept = −

9 2

From the equation 3x + 7y – 2 = 0, we have A = 3, B = 7 and C = –2.

2 7

Put y = 0 into 3x + 7y – 2 = 0, we have:

Alternative Solution y=

3 7

9.

∴

x-intercept =

∵

Slope = −

∴

10. ∵

2 3

A 3 3 A − =− 4 3 9 =4 A 9 A= 4

y-intercept =

−

(−6) B

26

12 Coordinate Treatment of Simple Locus Problems

∴

11. ∵ ∴

2 ( −6) =− 3 B 2 B =18 B =9 x-intercept = −

C 2

C −4 = − 2 C =8

C B y-intercept  −6  = −   3  =2 =−

14. 2(x – 3y) = x – 4y + 2 2x – 6y = x – 4y + 2 x – 2y – 2 = 0

Level 2 12.

∴

2(y + 3) = 3(x – 2) 2y + 6 = 3x – 6 3x – 2y – 12 = 0

A B 3  Slope = −    −2  3 = 2 =−

∴

C A x-intercept  −2  = −   1  =2 =−

C B y-intercept  −2  = −   −2  = −1 =−

C A x-intercept  −12  = −   3  =4 =−

C B y-intercept  −12  = −   −2  = −6 =−

y x + =1 2 3

13.

3y + 2x = 6 2x + 3y – 6 = 0 ∴

A B Slope 2 =− 3 =−

C A x-intercept  −6  = −   2  =3 =−

A B  Slope = − 1    −2  1 = 2 =−

2y −2 2 =− 3 x +1 3

15.

6y – 6 = –6x – 2 6x + 6y – 4 = 0 3x + 3y – 2 = 0

A B Slope 3 =− 3 = −1 =−

∴

C A −2  x-intercept = −    3  2 = 3 =−

27

Certificate Mathematics in Action Full Solutions 5A

C B  y-intercept = − − 2    3  2 = 3 =−

and −

b=−

C 3

 2   =1  −2 

20. (a) Let (0, a) and (b, 0) be the coordinates of A and B respectively.

y-intercept = −

−

C =1 3 C = −3

∵ ∴ ∵

17. For the straight line L1: 2x + 5y + 6 = 0, slope = −

2 5

∴

 8  = − =2  −4 

b = –5

 15    −5 

y-intercept = −

(b) y-intercept of L2 = y-intercept of L1 = 3 ∴

2 The equation of L is y = − x + 2. 5

The equation of L2 is y = −

(c) Put y = 0 into y = −

18. For the straight line L1: kx + 2y + 6 = 0, slope = −

15 3

x-intercept = −

∴ a=3 ∴ The coordinates of A and B are (0, 3) and (–5, 0) respectively.

For the straight line L2: x – 4y + 8 = 0, y-intercept

k 2

0=−

For the straight line L2: x + (k – 1)y + 3 = 0, slope = − ∴

1 k −1 k 1 − =− 2 k −1

k(k – 1) = 2 k2 – k – 2 = 0 (k + 1)(k – 2) = 0 1 or k= −

3 into (1), we have: 2

 3 a = −3−   2 9 = 2

For the straight line L2: 3x – 2y + 2 = 0,

∴

3 2

By substituting b = −

16. For the straight line L1: x + 3y + C = 0, y-intercept = −

6 =4 b

x=

=−

a b

y-intercept = −

9 2

∴

The coordinates of C = (

∴

Area of △ABC

k= 2

∴

Exercise 12F (p. 63)

6 b

a − =3 b a = –3b ……(1)

2 x + 3, we have: 3

2 x+3 3

19. For the straight line L: ax + by + 6 = 0, slope

2 x + 3. 3

9 , 0). 2

1 × BC ×OA 2 1 9  = × −( −5)  ×3 2 2  57 = 4 =

Level 1 1.

From the equation of L: x + y = 0, we have:

1

Slope of L = − 1

= −1 28

12 Coordinate Treatment of Simple Locus Problems

∵ ∴ ∴

2.

The required straight line // L. Slope of the required straight line = –1 The equation of the required straight line is y – 4 = –1(x – 3) y = –x + 7

∵ ∴

L =− =

1 ( −2)

1 2

∴

The required straight line // L.

∴

Slope of the required straight line =

∴

The equation of the required straight line is

y=

1 2

1 [x – (–2)] 2

7.

1

Slope of L = −1

= −1 ∵ ∴

2

=2 The required straight line // L. Slope of the required straight line = 2 The equation of the required straight line is y – 0 = 2(x – 1) y = 2x – 2

∴

8.

Slope of L

2 3

=−

2 ( −1)

=2

∵

The required straight line // L.

∴

Slope of the required straight line = −

∴

The equation of the required straight line is

∵ ∴

2 3

2 [x – (–5)] 3 2 10 y=− x− 3 3

From the equation of L: x = –3, we have: L // y-axis ∵ The required straight line // L. ∴ The required straight line // y-axis ∴ The equation of the required straight line is x = 2. Let m be the slope of the required straight line. From the equation of L: 2x + 3y = 0, we have:

The required straight line ⊥ L. m × slope of L = –1 m × 2 = –1 m=−

y–0=−

6.

The required straight line ⊥ L. m × slope of L = –1 m × (–1) = –1 m=1 The equation of the required straight line is y – 4 = 1(x – 0) y=x+4

Let m be the slope of the required straight line. From the equation of L: 2x – y – 7 = 0, we have:

From the equation of L: 2x + 3y – 4 = 0, we have: Slope of L = −

5.

Let m be the slope of the required straight line. From the equation of L: x + y – 5 = 0, we have:

1 x–2 2

Slope of L = − ( −1)

4.

The equation of the required straight line is

3 (x – 1) 2 3 7 x− y= 2 2

From the equation of L: 2x – y + 3 = 0, we have:

∵ ∴ ∴

The required straight line ⊥ L. m × slope of L = –1

y – (–2) =

∵

y – (–3) =

2 3

 2 m ×  −  = −1  3 3 m= 2

From the equation of L: x – 2y = 0, we have: Slope of

3.

Slope of L = −

∴

1 2

The equation of the required straight line is

1 y − 0 = − [ x − ( −2)] 2 1 y = − x −1 2 9.

Let m be the slope of the required straight line. From the equation of L: 5x – 4y – 6 = 0, we have: Slope of

L =− =

5 ( −4)

5 4

29

Certificate Mathematics in Action Full Solutions 5A

∵ ∴

The required straight line ⊥ L. m × slope of L = –1 m×

∴

5 = –1 4 4 m=− 5

4 = –1 3 3 m=− 4

m×

∴

The equation of L is

3 (x – 2) 4 3 11 y=− x+ 4 2

y–4=−

The equation of the required straight line is

4 [x – (–2)] 5 4 28 y=− x− 5 5

y – (–4) = −

14. ∵ ∴

L1 // L2 Slope of L1 = slope of L2

a a +2 =− 3 ( −2) 2a = −3a − 6

−

10. From the equation of L: y = 2, we have: L // x-axis ∵ The required straight line ⊥ L. ∴ The required straight line ⊥ x-axis i.e. The required straight line // y-axis ∴ The equation of the required straight line is x = 2. 11. From the equation of L1: 2x – y – 3 = 0, we have:

2 =2 Slope of L1 = − (−1) ∵ ∴ ∴

L // L1 Slope of L = 2 The equation of L is y – 2 = 2(x – 0) y = 2x + 2

12. Let m be the slope of L. From the equation of L1: 2x + 3y – 5 = 0, we have: Slope of L1 = − ∵ ∴

2 3

The equation of L is

3 y–0= (x – 0) 2 3 y= x 2 13. Let m be the slope of L. L1 passes through the points (–1, 0) and (2, 4).

4 −0 4 = Slope of L1 = 2 −( −1) 3 ∵ ∴

15. ∵ ∴

6 5

L1 ⊥ L2 Slope of L1 × slope of L2 = –1

−

a

( − 2)

 4 ×  −  = −1  5 5 a= 2

16. Let y = mx + c and y = nx + d be the equations of L1 and L2 respectively. ∵ L1 ⊥ L2 ∴ Slope of L1 × slope of L2 = –1 m × n = –1

L ⊥ L1 m × slope of L1 = –1

 2 m × −  = −1  3 3 m= 2 ∴

a =−

L ⊥ L1 m × slope of L1 = –1

n=−

1 m

∵ ∴ ∴

L1 intersects L2 at the y-axis. L1 and L2 have the same y-intercept. c=d

∴

L1 : y = mx + c and L2 : y = −

1 x+c m

∵

L1 has a positive x-intercept and L2 has a negative xintercept. Put y = 0 into the equations of L1 and L2, we have:

− ∴

c > 0 and cm < 0 m

The possible equations of L1 and L2 are y = x − 4 and y = −x − 4 or y = 2 x − 6 and

y =−

1 x − 6 respectively (or any other 2

reasonable answers). 17. For AB: y = 2x – 8, slope = 2 Put y = 0 into y = 2x – 8, we have:

30

12 Coordinate Treatment of Simple Locus Problems

0 = 2x – 8 x=4 ∴ x-intercept = 4 (a) ∵ BC ⊥ AB ∴ Slope of BC × slope of AB = –1 Slope of BC × 2 = –1 Slope of BC = − ∴

(c) ∵

Slope = −

1 2

1 1 x +10 or y = − x +12 (or 2 2

any other reasonable answers). CD // AB Slope of CD = slope of AB = 2 The possible equation of CD is y = 2 x − 2 or y = 2 x + 2 (or any other reasonable answers).

1 2

∴

Slope of AD = slope of BC = −

∵ ∴ ∴

AD cuts the x-axis at A. x-intercept of AD = x-intercept of AB = 4 The equation of AD is

1 (x – 4) 2 1 y=− x+2 2

Slope = −

 1 − 3 ×  −  = −1  a 3 a= −

∵ ∴

x-intercept of L2 = −

20. For L1: 4x + 3y – 6 = 0, Slope = −

slope of L2 =

∵

1 2

1  3   x −  −  2  2  1 3 y= x+ 2 4

19. For L1: 3x + y – 5 = 0,

5 3

The equation of L is y = −

Slope =

4 5 x− . 3 3

3 −( −1) 2 = 2 −( −4) 3

∵

L is perpendicular to the above straight line.

∴

Slope of L ×

∴

The equation of L is

The equation of L2 is

y −0=

5  −5  = − =− 3  −3  L // L1 L = slope of L1 Slope of 4 =− 3

y-intercept

21. For the straight line passing through B(–4, –1) and C(2, 3),

∵ L1 ⊥ L2 ∴ Slope of L1 × slope of L2 = –1 –2 × slope of L2 = –1

∴

4 3

For L2: 2x – 3y – 5 = 0,

∴

3 2

b 3

b=9

y-intercept of L = −

Slope = −

∴

L1 // L3 Slope of L1 = slope of L3 –3 = −

18. For L1: 2x + y + 3 = 0,

2 = –2 1 3 x-intercept = − 2

b 3

∵ L1 ⊥ L2 ∴ Slope of L1 × slope of L2 = –1

∴

Level 2

1 a

For L3: bx + 3y – 28 = 0,

AD // BC

y–0=−

3 = –3 1

For L2: x + ay – 6 = 0,

The possible equation of BC is

y =− (b) ∵ ∴ ∴

Slope = −

2 = –1 3 3 Slope of L = − 2

3 [x – (–2)] 2 3 y=− x+1 2

y–4=−

31

Certificate Mathematics in Action Full Solutions 5A

24. (a) Slope of OP =

−2 − 0 1 =− 22. (a) Slope of BC = 5 −1 2 ∵

AD // BC

∴

Slope of AD = slope of BC = −

∴

The required equation is

∵ ∴

1 2

1 [x – (–1)] 2 1 7 y=− x+ 2 2

∵ ∴ ∴

∵ ∴ ∴

23. (a) ∵ ∴

2 [x – (–2)] 3 2 13 y= x+ 3 3

y–3=

4 −0 = –2 −1 −1

(b) Rewrite the equation L: y =

x-intercept = −

4 −( −2) = –1 −1 −5

∴

∴

The equation of RQ is

1 y − 2 = − ( x − 8) 3 1 14 y =− x+ 3 3

1   13  39 × 0 −  −  × 3 = 2   2  4

(−8) =8 1 (−8) y-intercept = − =4 2 ∴

−

The coordinates of A and B are (8, 0) and (0, 4) respectively.

 8 +0 0 +4  ,  = (4, 2) 2   2

(b) K =  ∵ ∴

L1 ⊥ L2 Slope of L1 × slope of L2 = –1

−

1 × slope of L2 = –1 2

Slope of L2 = 2 The equation of L2 is y – 2 = 2(x – 4) y = 2x – 6

RQ // OP Slope of RQ = slope of OP = −

Area of △OAP =

x-intercept =

PQ // OR Slope of PQ = slope of OR = 3 The equation of PQ is y – 2 = 3(x – 8) y = 3x – 22

∴

13 2

25. (a) For the equation L1: x + 2y – 8 = 0,

OR ⊥ OP Slope of OR × slope of OP = –1 3m = –1 1 m=− 3

(ii) ∵

2 13 x+ into the 3 3

general form, we have 2x – 3y + 13 = 0.

BD ⊥ AC (property of rhombus) Slope of BD × slope of AC = –1 Slope of BD × (–1) = –1 Slope of BD = 1 The required equation is y – 0 = 1(x – 1) y=x–1

(b) (i) ∵ ∴ ∴

3 × slope of AB = –1 2 2 Slope of AB = 3

The equation of L is

DC // AB Slope of DC = slope of AB = –2 The required equation is y – (–2) = –2(x – 5) y = –2x + 8

(c) Slope of AC =

OP ⊥ AB Slope of OP × slope of AB = –1

−

y–4=−

(b) Slope of AB =

3 −0 3 =− −2 −0 2

1 3

(c) Rewrite the equation L2: y = 2x – 6 into the general form, we have 2x – y – 6 = 0.

(−6) =3 2 1 Area of △AKC = × (8 – 3) × 2 2

x-intercept = − ∴

=5

32

12 Coordinate Treatment of Simple Locus Problems

L2: 4y + 3 = 0 ……(2) From (2), we have:

Exercise 12G (p. 67) Level 1 1.

2.

∴



3x +  −



3 into (1), we have: 4

3 –3=0 4 x=

3 2

∴

 3  The coordinates of A = − 2 , −3   

L1: x – y – 1 = 0 L2: x + y – 3 = 0 (1) + (2),

3 4

By substituting y = −

L1: 2x – 3y – 6 = 0 ……(1) L2: y = –3 ……(2) By substituting (2) into (1), we have: 2x – 3(–3) – 6 = 0 x=−

3.

y=−

L1: x = –3 L2: y = 2 3, 2) ∴ The coordinates of A =(−

5 4

3 5 The coordinates of A =  4 , − 4   

3 −0 4 (b) Slope of L = 5 −0 4 −

……(1) ……(2)

( x − y −1) + ( x + y − 3) = 0 2x − 4 = 0

=−

The equation of L is y = −

x =2

4.

By substituting x = 2 into (1), we have: 2–y–1=0 y=1 ∴ The coordinates of A =( 2, 1) L1: 3x – 2y = 6 ……(1) L2: y = –x ……(2) By substituting (2) into (1), we have: 3x – 2(–x) = 6 x=

6 5

By substituting x = y=− ∴ 5.

6 into (2), we have: 5

6 5

8.

(a) L1: 2x – 5y + 1 = 0 ……(1) L2: x – 7 = 0 ……(2) From (2), we have: x=7 By substituting x = 7 into (1), we have: 2(7) – 5y + 1 = 0 y=3 ∴ The coordinates of A =(7, 3)

3 x. 7

(a) L1: 3x + y – 3 = 0 ……(1)

3 x. 5

(a) L1: 2x + y = –1 ……(1) L2: x – 2y = –8 ……(2) (1) – (2) × 2, (2x + y) – 2(x – 2y) = –1 – 2(–8) 5y = 15 y=3 By substituting y = 3 into (1), we have: 2x + 3 = –1 x = –2 2, 3) ∴ The coordinates of A =(− (b) The equation of L is y = −

6 6 The coordinates of A =  5 ,− 5   

(b) The equation of L is y = 6.

7.

3 5

3 x. 2

(a) L1: x – y = –1 ……(1) L2: 3x – y = 1 ……(2) (2) – (1), (3x – y) – (x – y) = 1 – (–1) 2x = 2 x=1 By substituting x = 1 into (1), we have: 1 −y = –1 y=2 ∴ The coordinates of A =(1, 2) (b) The equation of L is y = 2x.

Level 2 9.

L1: y = 2x – 2 ……(1) L2: y = –x + 4 ……(2) By substituting (2) into (1), we have:

33

Certificate Mathematics in Action Full Solutions 5A

–x + 4 = 2x – 2 x=2 By substituting x = 2 into (1), we have: y = 2(2) – 2 =2 ∴ The coordinates of A =( 2, 2)

0 − ( −2) 2 = 3 −0 3 2 L1: y – (–2) = (x – 0) 3 2 y= x – 2 ……(1) 3

10. Slope of L1 =

L2: y = 2x – 3 ……(2) By substituting (2) into (1), we have:

2 x–2 3 3 x= 4 3 By substituting x = into (2), we have: 4 3 y=2  –3 4 2x – 3 =

=− ∴

3 2

3 3 The coordinates of A =  4 ,− 2   

11. L1: y =

4 x 3

……(1)

4 x = –2x + 2 3 3 x= 5 3 By substituting x = into (1), we have: 5 4 3 y=   3 5

∴

∵ ∴

4 5

3 4  The coordinates of A =  5 , 5   

L2 ⊥ L1 Slope of L2 × slope of L1 = –1 Slope of L2 ×

 =2–1 −   33

Slope of L2 =

3 L2: y = x 2

2

……(2)

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

2 3 x=− x–2 3 2 12 x=− 13 12 By substituting x = − into (2), we have: 13 3  12  y = −  2  13  18 =− 13 18   12 ∴

L2: y = –2x + 2 ……(2) By substituting (1) into (2), we have:

=

0 − (−2) 2 =− −3 − 0 3 2 L1: y – (–2) = − (x – 0) 3 2 y = − x – 2 ……(1) 3

12. Slope of L1 =

−

The coordinates of A =  13

,−  13 

13. (a) L1: 2x – y – 3 = 0 ……(1) L2: x + y – 3 = 0 ……(2) (1) + (2), (2x – y – 3) + (x + y – 3) = 0 3x – 6 = 0 x=2 By substituting x = 2 into (2), we have: 2+y–3=0 y=1 ∴ The coordinates of P =( 2, 1) (b) ∵ ∴

L is parallel to the x-axis. The equation of L is y = 1.

14. (a) L1: 3x – y – 2 = 0 ……(1) L2: x – y – 4 = 0 ……(2) (1) – (2), (3x – y – 2) – (x – y – 4) = 0 2x + 2 = 0 x = –1 By substituting x = –1 into (2), we have: –1 – y – 4 = 0

34

12 Coordinate Treatment of Simple Locus Problems

∴

y = –5 1, − 5) The coordinates of A =( −

(b) Slope of L = ∴

0 − (−5) 5 = 2 − ( −1) 3

The equation of L is

5 (x – 2) 3 5 10 y= x− 3 3

y–0=

15. (a) Slope of L1 =

3 −0 =1 0 −( −3)

The equation of L1 is y – 3 = 1(x – 0) y = x + 3 ……(1) The equation of L2 is y = –3x – 3 ……(2) By substituting (1) into (2), we have: x + 3 = –3x – 3 x=−

3 2

= ∴

Slope of

3 into (1), we have: 2

3 2

∵ ∴

∴

17. (a) ∵ ∴

 3 3 The coordinates of P = − 2 , 2    3 2 3 − 2

−1

y=−

1 ( −2)

1 2

L ⊥ L3 Slope of L × slope of L3 = –1

−0

=−

1 3

1 (x – 0) 3

1 x+1 3

16. L1: (a + 1)x – 2y – 4 = 0 L2: 4x + (a + 3)y = 0 (a) ∵ L1 ⊥ L2 ∴ Slope of L1 × slope of L2 = –1

 a +1   4  −  × −  = −1  ( −2)   a + 3  2(a + 1) = a + 3 a =1

1 = –1 2

Slope of L = –2 The equation of L is y – (–1) = –2(x – 1) y = –2x + 1 L1 ⊥ L2 Slope of L1 × slope of L2 = –1 –2 × slope of L2 = –1 Slope of L2 =

∴

The equation of L is y–1=−

L3 = − =

3 +3 2

(b) Slope of L = − ∴

(c) From L3: x – 2y = 0, we have:

Slope of L ×

By substituting x = −

y =−

(b) From (a), we have: L1: (1 + 1)x – 2y – 4 = 0 x–y–2=0 ……(1) and L2: 4x + (1 + 3)y = 0 x+y=0 ……(2) (1) + (2), (x – y – 2) + (x + y) = 0 2x – 2 = 0 x=1 By substituting x = 1 into (2), we have: 1+y=0 y = –1 1) ∴ The coordinates of P =(1, −

1 2

The equation of L2 is

1 (x – 0) 2 1 y= x–4 2 1 Put y = 0 into y = x – 4, we have: 2 1 0= x–4 2 y – (–4) =

x=8 x-intercept of L2 = 8 Notice that L1 and L2 have the same x-intercept. ∴ x-intercept of L1 = 8 ∴ The equation of L1 is y – 0 = –2(x – 8) y = –2x + 16 ∵ L3 // L2 ∴ Slope of L3 = slope of L2

35

Certificate Mathematics in Action Full Solutions 5A

= ∴

1 2

2

The equation of L3 is y =

(b) From (a), we have: L1: y = –2x + 16 and L3: y =

1 x 2

     1  3 x −   + 3 y −  − 3      

1 x. 2

……(2)

 32 16  The coordinates of Q =  5 , 5   

Exercise 12H (p. 79) Level 1 1.

(a) The equation of the circle is (x – 0)2 + (y – 2)2 = 42 ∴ x2 + (y – 2)2 = 16 (b) The equation of the circle is (x – 1)2 + [y – (–1)]2 = 32 ∴ (x – 1)2 + (y + 1)2 = 9

2.

(a)

x2 + y2 = 49 (x – 0) + (y – 0)2 = 72 ∴ Centre = (0,0) and radius = 7 2

(b)

(c)

(x + 5)2 + (y – 2)2 = 9 [x – (–5)]2 + (y – 2)2 = 32 5,2) and radius = 3 ∴ Centre = (− (3x – 1)2 + (3y + 2)2 = 54 (3x – 1)2 + [3y – (–2)]2 = 54

2

2

2

2

1    2    x −  +  y −  −  = 6 3    3   1    2    x −  +  y −  −  = ( 6 ) 2 3    3  

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

∴

2

2

  2  1  9 x −  + 9 y −  −   = 54 3    3 

……(1)

1 x = –2x + 16 2 32 x= 5 32 By substituting x = into (2), we have: 5 1  32  y=   2 5  16 = 5

2

2       = 54 3   

∴

3.

Centre = (

1 2 ,− ) and radius = 3 3

6

10   ( −4) − ,−  (a) Centre =  2 2   = ( 2,−5) 2

2

−4   10   +   − 28 Radius =   2   2  =1

9  ( −6) =− ,−  2 2  (b) Centre 9  =3,−  2  2

2

−6  9 =   +  −0  2  2 Radius =

3 13 2

(c) 2x2 + 2y2 + 8y – 3 = 0 x2 + y2 + 4y –

3 =0 2

4  0 − ,−  Centre = 2 2  =(0, −2) 2

2

0 4  3 =   +   − −  2 2  2 Radius =

22 2

36

12 Coordinate Treatment of Simple Locus Problems

4.

5.

6.

As the circle touches the y-axis, its radius is 5. The equation of the circle is [x – (–5)]2 + (y – 3)2 = 52 x2 + 10x + 25 + y2 – 6y + 9 = 25 x2 + y2 + 10x – 6y + 9 = 0 As the circle touches the x-axis, its radius is 3. The equation of the circle is [x – (–1)]2 + [y – (–3)]2 = 32 2 x + 2x + 1 + y2 + 6y + 9 = 9 x2 + y2 + 2x + 6y + 1 = 0 2 2 (a) Radius = (3 − 0) + ( 4 − 0) =5 The equation of the circle is x2 + y2 = 52 x2 + y2 – 25 = 0

(b) Radius = (5 −1) 2 +[6 −( −2)] 2

=4 5 The equation of the circle is (x – 1)2 + [y – (–2)]2 = ( 4 5 ) 2 x2 + y2 – 2x + 4y – 75 = 0 7.

For the circle C: x2 + y2 – 4x + 8y – 7 = 0, 2

2

−4 8  =   +   − ( −7) radius  2  2 =3 3 Area of the circle

=π(3 3 ) 2 =27 π

8.

∵

9.

2 2 (a) Radius = (1 −3) +( 4 −2)

The circle touches the positive x-axis and the positive y-axis. ∴ Its centre is (3, 3). The equation of the circle is (x – 3)2 + (y – 3)2 = 32 2 x + y2 – 6x – 6y + 9 = 0

=2 2

The equation of the circle is (x – 3)2 + (y – 2)2 = ( 2 2 ) 2 x2 + y2 – 6x – 4y + 5 = 0 (b) By substituting y = 0 into x2 + y2 – 6x – 4y + 5 = 0, we have: x2 + 02 – 6x – 4(0) + 5 = 0 x2 – 6x + 5 = 0 (x – 1)(x – 5) = 0 x = 1 or x = 5 ∴ The coordinates of the points where the circle cuts the x-axis are (1, 0) and (5, 0).

2 2 10. (a) Radius = ( 2 −1) +[3 −( −2)]

=

26

The equation of the circle is (x – 1)2 + [y – (–2)]2 = ( 26 ) 2 2 x + y2 – 2x + 4y – 21 = 0 (b) By substituting x = 0 into x2 + y2 – 2x + 4y – 21 = 0, we have: 02 + y2 – 2(0) + 4y – 21 = 0 y2 + 4y – 21 = 0 (y – 3)(y + 7) = 0 y=3 or y = –7 ∴ The coordinates of the points where the circle cuts the y-axis are (0, 3) and (0, –7). 11. (a) L: 3x – 2y + 6 = 0 By substituting y = 0 into the equation 3 x − 2 y + 6 = 0 , we have: 3x – 2(0) + 6 = 0 x = –2 By substituting x = 0 into the equation 3 x − 2 y + 6 = 0 , we have: 3(0) – 2y + 6 = 0 y=3 ∴ The coordinates of A and B are (–2, 0) and (0, 3) respectively. (b) Let the equation of the circle be: x2 + y2 + Dx + Ey + F = 0 ……(1) By substituting (–2, 0) into (1), we have: (–2)2 + 02 + D(–2) + E(0) + F = 0 i.e. –2D + F = –4 ……(2) By substituting (0, 3) into (1), we have: 02 + 32 + D(0) + E(3) + F = 0 i.e. 3E + F = –9 ……(3) By substituting (0, 0) into (1), we have: 02 + 02 + D(0) + E(0) + F = 0 i.e. F=0 By substituting F = 0 into (2), we have: –2D + 0 = –4 D=2 By substituting F = 0 into (3), we have: 3E + 0 = –9 E = –3 ∴ The equation of the circle is x2 + y2 + 2x – 3y = 0.

Level 2 12. For the circle C: x2 + y2 – 6x + 4y – 4 = 0,

( −6) 4 centre =  ,−  − 2  = (3, −2)

2

37

Certificate Mathematics in Action Full Solutions 5A

2

2

 −6  4 =   +   − ( −4) radius  2  2

∴

= 17 = (8 −3) 2 +[5 −( −2)] 2 = 74 > 17

The point (8, 5) lies outside the circle C.

13. By substituting (2, –5) into the equation of C: (x + a)2 + (y + 3)2 = a + 8, we have: (2 + a)2 + (–5 + 3)2 = a + 8 4 + 4a + a2 + 4 = a + 8 a2 + 3a = 0 a(a + 3) = 0 3 a=0 or a = −

2

2

2

D E  −2   4    +  − F =   +   −8 2 2  2  2 = −3 ∴

<0 C: x2 + y2 – 2x + 4y + 8 = 0 represents an imaginary circle.

15. For the circle C: x2 + y2 + 2x + 6y – 2k = 0, 2

2

2

2

D E 2 6   +   − F =   +   − (−2k ) 2 2     2 2 = 10 + 2k If C represents a real circle or a point circle, 2

2

D E   +  − F ≥ 0 2 2 ∴

∴

10 + 2k ≥ 0 2k ≥ –10 k ≥ –5 The range of values of k is k ≥ –5.

16. For the circle C: x2 + y2 – 2x + 4y – 4 = 0,

4  ( −2) ,−  = (1, –2) centre =  − 2 2  By substituting (1, –2) into the equation of L: 2x + 3y + b = 0, we have: 2(1) + 3(–2) + b = 0 b= 4 17. Let the equation of the circle be x2 + y2 + Dx + Ey + F = 0.



Centre =  −



∵

E=0 The equation of the circle becomes x2 + y2 + Dx + F = 0 ……(1) By substituting (4, 2) into (1), we have: 42 + 22 + D(4) + F = 0 i.e. 4D + F = –20 ……(2) By substituting (–6, –2) into (1), we have: (–6)2 + (–2)2 + D(–6) + F = 0 i.e. –6D + F = –40 ……(3) (2) – (3), 10D = 20 D=2 By substituting D = 2 into (2), we have: 4(2) + F = –20 F = –28 ∴ The equation of the circle is x2 + y2 + 2x – 28 = 0. 18. Let the equation of the circle be x2 + y2 + Dx + Ey + F = 0.

14. For the circle C: x2 + y2 – 2x + 4y + 8 = 0, 2

E =0 2

∴

Distance between the point (8, 5) and the centre

∴

−

D E ,−  2 2

The centre lies on the x-axis.

 

Centre =  −

D E ,−  2 2

∵

The centre lies on the y-axis.

∴

−

D =0 2

D=0 The equation of the circle becomes x2 + y2 + Ey + F = 0 ……(1) By substituting (–1, 1) into (1), we have: (–1)2 + 12 + E(1) + F = 0 i.e. E + F = –2 ……(2) By substituting (−2, 0) into (1), we have: (–2)2 + 02 + E(0) + F = 0 i.e. F = –4 By substituting F = –4 into (2), we have: E + (−4) = –2 E=2 ∴ The equation of the circle is x2 + y2 + 2y – 4 = 0. ∴

19. Let the equation of the circle be x2 + y2 + Dx + Ey + F = 0.

 

Centre =  −

D E ,−  2 2

∵

The centre lies on the straight line x + y = 0.

∴

−

D  E + −  = 0 2  2

D = –E The equation of the circle becomes x2 + y2 – Ex + Ey + F = 0 ……(1) By substituting (1, 2) into (1), we have: 12 + 22 – E(1) + E(2) + F = 0 i.e. F = –5 – E ……(2) By substituting (5, 0) into (1), we have: 52 + 02 – E(5) + E(0) + F = 0 i.e. –5E + F = –25 ……(3) By substituting (2) into (3), we have: ∴

38

12 Coordinate Treatment of Simple Locus Problems

–5E + (–5 – E) = –25

Slope of BC =

10 E= 3 ∴

Slope of AB × slope of BC = –2 ×

D = –E =−

10 3 10 into (2), we have: 3

10 3 25 =− 3

F = –5 –

∴ ∴

AB ⊥ BC ∠ ABC = 90°

(b) AC is the diameter of the circumcircle of ABC. (converse of ∠ in semi-circle)  2 +10 3 + ( −3)  ,  Centre of the circle = 2  2 

= ( 6,0 )

The equation of the circle is

x2 + y2 −

10 10 25 x+ y− =0 . 3 3 3

20. (a) By substituting x = 0 into the equation of C: x2 + y2 + 6x – 10y + 9 = 0, we have: 02 + y2 + 6(0) – 10y + 9 = 0 y2 – 10y + 9 = 0 (y – 1)(y – 9) = 0 y=1 or y = 9 ∴ The coordinates of A and B are (0, 9) and (0, 1) respectively. By substituting y = 0 into the equation of C: x2 + y2 + 6x – 10y + 9 = 0, we have: x2 + 02 + 6x – 10(0) + 9 = 0 x2 + 6x + 9 = 0 (x + 3)2 = 0 x = –3 ∴ The coordinates of P are (–3, 0).

(b) Slope of AP

(c) ∵ ∴

1 2

= –1

By substituting E =

∴

−3 −( −5) 1 = 10 −6 2

0 −9 −3 − 0 =3 =

L ⊥ AP Slope of L × slope of AP = –1 Slope of L × 3 = –1 Slope of L = −

The equation of L is

1 (x – 0) 3 1 y=− x+1 3

y–1=−

x + 3y − 3 = 0 21. (a) Slope of AB =

−5 −3 6 −2

1 3

2 2 Radius of the circle = ( 2 −6) +(3 −0)

=5

(c) The equation of the circle is (x – 6)2 + (y – 0)2 = 52 ∴ x2 + y2 – 12x + 11 = 0 22. (a) By substituting y = 0 into the equation of C: x2 + y2 + 10x – 8y + 16 = 0, we have: x2 + 02 + 10x – 8(0) + 16 = 0 x2 + 10x + 16 = 0 (x + 2)(x + 8) = 0 x = –2 or x = –8 ∴ The coordinates of A and B are (–8, 0) and 2, 0) respectively. By substituting x = 0 into the equation of C: x2 + y2 + 10x – 8y + 16 = 0, we have: 02 + y2 + 10(0) – 8y + 16 = 0 y2 – 8y + 16 = 0 (y – 4)2 = 0 y=4 ∴ The coordinates of P are (0, 4).

(b) Slope of BP

(–

4 −0 0 − (−2) =2 =

∵ ∴

AQ // BP Slope of AQ = slope of BP =2 The equation of AQ is y – 0 = 2[x – (–8)] y = 2x + 16 ……(1) 2x – y + 16 = 0 By substituting x = 0 into (1), we have: y = 2(0) + 16 = 16 ∴ The coordinates of Q are (0, 16). (c) Area of trapezium ABPQ = Area of AOQ – area of BOP

= –2

39

Certificate Mathematics in Action Full Solutions 5A

2 − ( −3) 8 − (−2) Slope of BC 1 = 2

1 1 [0 − ( −8)]( 16 − 0) − [0 − ( −2)]( 4 − 0) 2 2 = 60 =

=

Revision Exercise 12 (p. 88)

∵ ∴

Level 1 1.

(a) (i) Slope of AB

Slope of BC

5 −0 3 −( −2) =1 =

3.

(a) K is the mid-point of A and C. ∴ The coordinates of K

9 −5 7 −3 =1 =

(ii) Let θ1 and θ2 be the inclinations of AB and BC respectively. tanθ1 = 1 θ1 = 45° tanθ2 = 1 θ2 = 45° ∴ The inclinations of AB and BC are 45° and 45° respectively.

∴ ∴ ∴ 4.

x +5 y +6 = 4 and =8 2 2 x = 3 and y = 10 The coordinates of D =(3, 10 )

L : 2x − 3 y + 5 = 0

3y = 2x + 5

1

2 5 y = x+ 3 3 2 Slope of L1 = 3 ∵ ∴

L // L1 Slope of L = slope of L1 =

(a) The coordinates of

 3 + ( −2) 4 + ( −3 )  M = ,  2 2   1 1   = ,  2 2   3 +8 4 + 2  N = ,  2   2 The coordinates of  11  =  ,3  2 

= (b) Slope of MN

=

3− 11 2

1 2

−

1 2 1 2

1 +7 4 +12  = ,  2   2 = ( 4,8)

(b) Let (x, y) be the coordinates of D. ∵ K is the mid-point of B and D.

(b) ∵ Slope of AB = slope of BC ∴ AB // BC ∵ AB and BC contain the common point B. ∴ ABC is a straight line. ∴ A, B and C are collinear. 2.

Slope of MN = slope of BC MN // BC

2 3

The equation of L is y=

2 x 3

2x – 3y = 0 5.

For the straight line L: y = mx + 4, slope = m. (a) For the straight line L1: (2m + 1)x + y = 0, slope = –(2m + 1) ∵ L // L1 ∴ Slope of L = slope of L1 m = –(2m + 1) 1 m= − 3 (b) For the straight line L2: x + (3m + 1)y – 4 = 0, slope = − ∵ ∴

1 3m +1

L ⊥ L2 Slope of L × slope of L2 = –1

1   m ×−  = −1  3m + 1 

40

12 Coordinate Treatment of Simple Locus Problems

m= − 6.

∵ ∴

Slope of AP = slope of BP

1 2

L ⊥ L1 Slope of L × slope of L1 = –1 Slope of L ×

 =3–1 −   5 5

Slope of L =

(b) Let (0, y) be the coordinates of Q. ∵ B, P and Q are on the same straight line. ∴ Slope of BP = slope of PQ

3

The equation of L is y=

5 x+3 3

5x – 3y + 9 = 0 7.

By substituting y = 0 into the equation 2x – y + 8 = 0, we have: 2x – 0 + 8 = 0 x = –4 ∴ The coordinates of A = (–4, 0) ∵ L ⊥ L1 ∴ Slope of L × slope of L1 = –1

The equation of L is y–0=−

1 [x – (–4)] 2

x + 2y + 4 = 0 (a) L1: x + y = 7 ……(1) L2: x – y = 3 ……(2) By substituting (a, b) into (1), we have: a + b = 7 ……(3) By substituting (b, a) into (2), we have: b – a = 3 ……(4) (3) + (4), 2b = 10 b=5 By substituting b = 5 into (3), we have: a+5=7 a= 2

2 −5 (b) Slope of PQ = 5 −2 = –1 The required equation is y – 5 = –1(x – 2) x+y–7=0 9.

(a) ∵

∴

0 −( −4) y −0 = 1 −( −3) 0 −1 (0, − 1) −1 The coordinatesyof = Q =

10. 3a = 4b

2  Slope of L ×  = –1  − ( −1)     1 Slope of L = − 2

8.

p −0 0 − ( −4) = 4 −1 1 − ( −3) p =3

∴

A, B and P are on the same straight line.

b 3 = a 4 Slope of L =

b −0 b 3 =− =− 0 −a a 4

The equation of L is y–3=−

3 (x – 2) 4

3x + 4y – 18 = 0 11. Let a be the x-intercept and y-intercept of the straight line where a > 0. Slope of the straight line =

a −0 0 −a

= –1 The required equation is y − 4 = −1( x − 3)

x + y −7 = 0 12.

Let the perpendicular bisector be L. ∵ L ⊥ AB ∴ Slope of L × slope of AB = –1 Slope of L ×

12 −4 = –1 7 −3

Slope of L × 2 = –1 Slope of L = −

1 2

41

Certificate Mathematics in Action Full Solutions 5A

 3 + 7 4 +12  ,  2   2

−

Mid-point of AB = 

= (5, 8) ∵ L bisects AB. ∴ L passes through the mid-point of AB. The equation of L is y–8=−

1 (x – 5) 2

x + 2y – 21 = 0 13. (a) L1: x + y – 2 = 0 ……(1) L2: 2x + y – 4 = 0 ……(2) (2) – (1), (2x + y – 4) – (x + y – 2) = 0 x–2=0 x=2 By substituting x = 2 into (1), we have: 2+y–2=0 y=0 ∴ The coordinates of A =( 2, 0) (b) ∵ ∴

Slope of OB = –2 The required equation is y = –2x 2x + y = 0 ……(1) (c) L: x – 2y – 6 = 0 x = 2y + 6 ……(2) By substituting (2) into (1), we have:

2(2 y + 6) + y = 0 5 y + 12 = 0

y=−

The equation of L is y–0=

1 (x – 2) 2

x – 2y – 2 = 0 14. (a)

12 5

By substituting y = −

12 into (2), we have: 5

 12  +6  5  6 = 5

x = 2 −

L ⊥ L2 Slope of L × slope of L2 = –1

 2 Slope of L ×  −  = –1  1 1 Slope of L = 2

1 × slope of OB = –1 ( −2)

∴

12  6 The coordinates of B =  5 , − 5   

15. (a) The equation of the circle is x 2 + y 2 = 22 ∴ x2 + y2 −4 = 0 2 2 (b) Radius = (3 −0) +( −2 −1)

=3 2

The equation of the circle is (x – 0)2 + (y – 1)2 = (3 2 ) 2 ∴ x2 + y2 – 2y – 17 = 0

 −2 + 2 3 + ( −5)  ,  2  2 

(c) Centre = 

= (0, –1) Radius = Let θ be the inclination of L. Slope of L = tan θ

−

1 = tan θ ( −2)

θ = 26.57° ∠ OAB = θ (vert. opp. ∠ s) = 26 .6° (cor. to the nearest 0.1°) (b) ∵ ∴

L ⊥ OB Slope of L × slope of OB = –1

=

1 × diameter 2 1 × [ 2 − (−2)] 2 + ( −5 − 3) 2 2

=2 5 The equation of the circle is (x – 0)2 + [y – (–1)]2 = ( 2 5 ) 2 ∴ x2 + y2 + 2y – 19 = 0 (d) ∵ The circle touches the y-axis. ∴ Radius = 0 – (–2) = 2 The equation of the circle is [x – (–2)]2 + (y – 3)2 = 22

42

12 Coordinate Treatment of Simple Locus Problems

∴ x2 + y2 + 4x – 6y + 9 = 0

17  20  3 − = −13  + 2E + 3  3  2 E= 3

16. (a) The centre (0, k) is at the same distance from (3, 5) and (5, –1). ∴

(0 − 3) 2 + ( k − 5) 2 =

∴ The equation of the circle is

(0 − 5) 2 +[ k − ( −1)] 2

9 + k 2 −10 k + 25 = 25 + k 2 + 2k +1 12 k = 8 2 k = 3

2  = (0 − 3) 2 +  − 5  (b) Radius 3   5 = 10 3

2

x2 + y2 −

( ) ()

2  − 20  =  − 3 ,− 3   2 2    (b) Centre 10 1   = ,−  3  3

Radius =

The equation of the circle is 2

2  5  ( x − 0) 2 +  y −  =  10  3  3  ∴ 4 82 x2 + y2 − y − =0 3 3

2

17. (a) Let the equation of the circle be: x2 + y2 + Dx + Ey + F = 0 ……(1) By substituting (1, 0) into (1), we have: 12 + 02 + D(1) + E(0) + F = 0 i.e. D + F = –1 ……(2) By substituting (3, 2) into (1), we have: 32 + 22 + D(3) + E(2) + F = 0 i.e. 3D + 2E + F = –13 ……(3) By substituting (5, −2) into (1), we have: 52 + (–2)2 + D(5) + E(–2) + F = 0 i.e. 5D – 2E + F = –29 ……(4) (3) + (4), 8D + 2F = –42 4D + F = –21 ……(5) (5) – (2), 3D = –20 D=−

20 3

By substituting D = −

−

20 + F = –1 3 17 F= 3

By substituting D = − have:

20 into (2), we have: 3

20 17 and F = into (3), we 3 3

20 2 17 x+ y+ = 0. 3 3 3

=

 −320   2 

2

2

  23  17  +  −  2 3   

5 2 3

18. Let the equation of the circle be x2 + y2 + Dx + Ey + F = 0.

 

Centre =  −

D E ,−  2 2

∵

The centre lies on the line x + y = 0.

∴

−

D  E + −  = 0 2  2

D = –E The equation of the circle becomes x2 + y2 – Ex + Ey + F = 0 ……(1) By substituting (–3, 0) into (1), we have: (–3)2 + 02 – E(–3) + E(0) + F = 0 i.e. F = –9 – 3E ……(2) By substituting (0, 5) into (1), we have: 02 + 52 – E(0) + E(5) + F = 0 i.e. 5E + F = –25 ……(3) By substituting (2) into (3), we have: 5E + (–9 – 3E) = –25 E = –8 ∴ D = –E = 8 By substituting E = –8 into (2), we have: F = –9 –3(–8) = 15 ∴ The equation of the circle is x2 + y2 + 8x – 8y + 15 = 0. ∴

19. (a) For the equation C1: x2 + y2 + 4x + 6y – 10 = 0, 6  4 = − ,−  centre 2 2  = ( −2, −3) (b) ∵

C1 and C2 are concentric.

43

Certificate Mathematics in Action Full Solutions 5A

∴

Centre of C2 = centre of C1 = (–2, –3) Radius of C 2

= [5 −( −2)] 2 +[ −3 −( −3)] 2 =7 ∴ The equation of the circle is [x – (–2)]2 + [y – (–3)]2 = 72 x2 + y2 + 4x + 6y – 36 = 0 20. Let (x, 0) and (0, y) be the coordinates of A and B respectively. ∵ AB = 10 ∴

−

(−11 ) = 11 1

∵ ∴

L lies in the shaded region. x-intercept of L is less than –3 or greater than 11.

∴

−

4 −3m 4 − 3m < −3 or − >11 m m 2 1 m< or m <− 3 2

Let m = 0. ∴ A possible equation of L is

0 x − y +[4 −3(0)] = 0 y =4

( x −0) 2 +(0 −y ) 2 = 10 2

2

x + y = 100 Put x = 6, then y = 8.

0 −8 6 −0 Slope of L 4 =− 3

Let m = ∴

=

∴

x-intercept of AC =

A possible equation of L is y=−

4 x+8 3

4x + 3y – 24 = 0 Put x = −8 , then y = 6 .

1 . 2

A possible equation of L is

1 1  x − y + [4 − 3 ] = 0 2 2 x −2y +5 = 0 1 Let m = . 3 ∴

A possible equation of L is

Slope of L =

1 1  x − y + [ 4 − 3 ] = 0 3 3  x −3y +9 = 0 1 Let m = − . 3

∴

∴

6 −0 0 − ( −8) 3 = 4

A possible equation of L is

y= ∴

3 x +6 4

3x − 4 y + 24 = 0 The possible equation of straight line L is 4 x + 3 y − 24 = 0 or 3x − 4 y + 24 = 0 (or any other reasonable answers).

21. AB: 2x – 3y + 6 = 0 ……(1) AC: x + 2y – 11 = 0 ……(2) (2) × 2 – (1), 2(x + 2y – 11) – (2x – 3y + 6) = 0 y=4 By substituting y = 4 into (2), we have: x + 2(4) – 11 = 0 x=3 ∴ The coordinates of A are (3, 4). Let m be the slope of L. The equation of L is y – 4 = m(x – 3) mx – y + (4 – 3m) = 0 x-intercept of AB = −

A possible equation of L is

6 = –3 2

−

1  1 x − y + [ 4 − 3 − ] = 0 3  3 x + 3 y −15 = 0

∴ The two possible equations of straight line L are y = 4 , x − 2 y + 5 = 0 , x − 3 y + 9 = 0 or x + 3 y −15 = 0 (or any other reasonable answers).

Level 2 22. (a) Slope of AB = 2

y −0 =2 6 −5 y= 2 (b) ∵ ∴

L ⊥ AB Slope of L × slope of AB = –1 Slope of L × 2 = –1 Slope of L = −

1 2

The equation of L is

44

12 Coordinate Treatment of Simple Locus Problems

y–2=− x + 2y – 10 = 0

1 (x – 6) 2 ……(1)

L1 ⊥ L and L ⊥ AB L1 // AB Slope of L1 = slope of AB =2 The equation of L1 is y = 2x ……(2) By substituting (2) into (1), we have: x + 2(2x) – 10 = 0 x=2 By substituting x = 2 into (2), we have: y = 2(2) =4 ∴ The coordinates of intersection are (2, 4).

(c) ∵ ∴ ∴

23. x + y = 2 ……(1) x – y = –6……(2) (1) + (2), 2x = –4 x = –2 (1) – (2), 2y = 8 y=4 ∴ The coordinates of intersection are (–2, 4).

4 −0 − 2 −5 Slope of the required line 4 =− 7 =

The required equation is y–0=−

=−

4 (x – 5) 7

4x + 7y – 20 = 0

3 4

The required equation is y–1=−

3 [x – (–2)] 4

3x + 4y + 2 = 0

……(1)

(iv) Slope of CD = slope of AB =

4 3

The required equation is

4 y–2= (x – 5) 3 4x – 3y – 14 = 0

……(2)

(b) (1) × 3 + (2) × 4, 3(3x + 4y + 2) + 4(4x – 3y – 14) = 0 9x + 12y + 6 + 16x – 12y – 56 = 0 25x – 50 = 0 x=2 By substituting x = 2 into (1), we have: 3(2) + 4y + 2 = 0 y = –2 ∴ The coordinates of D =( 2, −2) (c) ∵ ∴

BD bisects AC. (property of square) K is the mid-point of AC.

 −2 + 5 1 + 2  = ,  2   2 The coordinates of K 3 3 = ,  2 2 25. (a) L: 2x + y – 4 = 0

24. (a) (i) Slope of AB =

5 −1 4 = 1 −( −2) 3

The required equation is y–1=

4 [x – (–2)] 3

(−4) =2 2 (−4) y-intercept = − =4 1 x-intercept = −

∴

4x – 3y + 11 = 0

2 −5 5 −1 (ii) Slope of BC 3 =− 4 =

The required equation is y–5=−

3 (x – 1) 4

3x + 4y – 23 = 0

The coordinates of A and B are (2, 0) and (0, 4) respectively.

(b) Slope of L = ∵ ∴

4 −0 0 −2

= –2 L ⊥ L1 Slope of L × slope of L1 = –1 –2 × slope of L1 = –1 Slope of L1 =

1 2

The equation of L1 is

(iii) Slope of AD = slope of BC

45

Certificate Mathematics in Action Full Solutions 5A

y–0=

1 (x – 2) 2

x – 2y – 2 = 0 ……(1) (c) ∵ ∴

L2 // x-axis. The equation of L2 is: y=4 ……(2) By substituting (2) into (1), we have x – 2(4) – 2 = 0 x = 10 ∴ The coordinates of C =(10 , 4)

26. (a) Slope of L2 = − ∵ ∴

1 2

L1 ⊥ L2 Slope of L1 × slope of L2 = –1

 =1–1 −   2 Slope of L = 2

= –3 The equation of straight line passing through A and D is y – 8 = –3(x – 6) 3x + y – 26 = 0 (d) From (b), x-intercept of y = kx + 6 is 2. ∴ The coordinates of B =( 2, 0) From the equation of straight line passing through A and D: 3x + y – 26 = 0,

( −26 ) 3 x-intercept 26 = 3 =−

∴

 26  The coordinates of D =  3 , 0   

Slope of L1 ×

1

The equation of L1 is y – 5 = 2(x – 5) 2x – y – 5 = 0 ……(1) (b) x + 2y – 5 = 0 x = 5 – 2y ……(2) By substituting (2) into (1), we have: 2(5 – 2y) – y – 5 = 0 y=1 By substituting y = 1 into (2), we have: x = 5 – 2(1) =3 ∴ The coordinates of intersection are (3, 1).

(e) Area of ABD = area of BCD ∴ Area of parallelogram ABCD = 2 × area of ABD

 1  26   = 2 ×  × − 2 ×8   2  3 160 = 3 28. (a) Let (0, y) be the coordinates of A. ∵ AO = AP

y = (6 − 0) 2 + ( 2 − y ) 2 ∴

4 y = 40 y =10

(c) The perpendicular distance from A to L2 = the distance between A and the point (3, 1) = (5 −3) 2 +(5 −1) 2 =2

27. (a) ∵ ∴

∴

The coordinates of A =(0, 10 )

5

A(6, 8) lies on the line y = mx – 4. 8 = m(6) – 4 m= 2

(b) The straight line passing through A and B is y = 2x – 4 2x – y – 4 = 0

(−4) x-intercept = − =2 2 x-intercept of line joining AB = x-intercept of line joining BC =2 ∴ 0 = k(2) + 6 3 k= − (c) ∵ ∴

y 2 = 36 + 4 − 4 y + y 2

AD // BC Slope of AD = slope of BC

10 − 2 0 −6 (b) Slope of AP 4 =− 3 =

∵ ∴

AP ⊥ PB Slope of AP × slope of PB = –1

−

4 × slope of PB = –1 3 3 Slope of PB = 4

The equation of the straight line passing through P and B is y–2=

3 (x – 6) 4

46

12 Coordinate Treatment of Simple Locus Problems

3x – 4y – 10 = 0

……(1) 30. (a) Centre

(c) In OAB and PAB, AB = AB AO = AP ∠ AOB = ∠ APB = 90° ∴ OAB ≅ PAB

common side given given RHS

(d) From the equation of straight line passing through P and B: 3x – 4y – 10 = 0, x-intercept ∴

=−

(−10 ) 10 = 3 3

10 The coordinates of B = ( , 0). 3

Area of quadrilateral OAPB = 2 × area of OAB

2

The distance between M and the centre of circle C = [1 −( −2)] 2 +( 4 −3) 2 = 10

<4 = radius of circle C ∴ M is a point inside the circle.

4 −3 1 − (−2) Slope of PM 1 = 3 =

 1  10   = 2 ×  × − 0 ×(10 −0)  2 3     100 = 3 2

(−6)   2 = − , −  2   2 = (−1, 3) 2

∵ ∴

PM ⊥ AB

(line joining centre to mid-pt. of chord ⊥ chord) Slope of PM × slope of AB = –1

1 × slope of AB = –1 3

2

29. (a) By substituting (4, 3) into x + y + 2x + ky – 15 = 0, we have: 42 + 32 + 2(4) + k(3) – 15 = 0 3k = –18 6 k= −

2

2  −6   − ( −15 ) Radius =   +  2  2  =5 The distance between B and the centre of circle C =

2

4  −6  − (−3) Radius =   +  2  2  =4

(b) Let P be the centre of the circle C.

1 =2× × OB × OA 2

(b) Centre

( −6)   4 = − , −  2   2 = ( −2, 3)

[ −1 −( −6)] 2 +(3 −3) 2

=5 = radius of circle C ∴ The point B lies on the circle.

 4 + ( −6) 3 + 3  ,  2 2  

(c) Mid-point of A and B = 

= (–1, 3) = centre of circle C Also, A and B lie on the circle. ∴ AB is a diameter of the circle C.

Slope of AB = –3 The equation of the chord AMB is y – 4 = –3(x – 1) 3x + y – 7 = 0 31. (a) The equation of L is y – 0 = 1[x – (–4)] x–y+4=0 (b) Let (x, y) be the coordinates of C. ∵ C lies on L. ∴ y=x+4 ∴ The coordinates of C = (x, x + 4). ∵ CO = CB ( x −0) 2 +[( x +4) −0] 2

∴ =

( x −8) 2 +[( x +4) −0] 2

x 2 + x 2 + 8 x +16 = x 2 −16 x + 64 + x 2 + 8 x +16 16 x = 64 x =4

By substituting x = 4 into y = x + 4 , we have:

y = 4 +4 =8

∴

The coordinates of C =( 4, 8)

(c) Let the equation of the circle be: x2 + y2 + Dx + Ey + F = 0 ……(1)

47

Certificate Mathematics in Action Full Solutions 5A

By substituting (8, 0) into (1), we have: 82 + 02 + D(8) + E(0) + F = 0 i.e. 8D + F = –64 ……(2) By substituting (4, 8) into (1), we have: 42 + 82 + D(4) + E(8) + F = 0 i.e. 4D + 8E + F = –80 ……(3) By substituting (0, 0) into (1), we have: 02 + 02 + D(0) + E(0) + F = 0 i.e. F=0 By substituting F = 0 into (2), we have: 8D + 0 = –64 D = –8 By substituting D = –8 and F = 0 into (3), we have: 4(–8) + 8E + 0 = –80 E = –6 ∴ The equation of the circle is x2 + y2 – 8x – 6y = 0. 2

2

 −8 −6  +  −0 (d) Radius =   2   2  =5 Area of the circle = π (52)

=78 .5 (cor. to 3 sig. fig.)

32. (a) From the equation of L: 3x + 4y – 24 = 0, x-intercept = −

=8 y-intercept = −

( −24 ) 3

(c) tan ∠ OBA

OA OB 8 = 6 =

∴ ∠ OBA = 53.13° ∠ OPA + ∠ OBA = 180° (opp. ∠ s, cyclic quad.) ∠ OPA + 53.13° = 180° ∠ OPA = 127 ° (cor. to the nearest degree) 33. Let the equation of the circles be (x – h)2 + (y – k)2 = r2. Since the circles touch the x-axis and y-axis, h = k = r. The equation of the circles become (x – r)2 + (y – r)2 = r2 x2 + y2 – 2rx – 2ry + r2 = 0 ……(1) By substituting (4, 2) into (1), we have: 42 + 22 – 2r(4) – 2r(2) + r2 = 0 r2 – 12r + 20 = 0 (r – 2)(r – 10) = 0 r = 2 or r = 10 ∴ The equation of the circle C1 is x2 + y2 – 2(2)x – 2(2)y + 22 = 0 x2 + y2 – 4x – 4y + 4 = 0. The equation of the circle C2 is x2 + y2 – 2(10)x – 2(10)y + 102 = 0 x2 + y2 – 20x – 20y + 100 = 0. 34.

( −24 ) 4

=6 ∴

The coordinates of A and B are (8, 0) and (0, 6) respectively.

(b) Let the equation of the circle be: x2 + y2 + Dx + Ey + F = 0 ……(1) By substituting (8, 0) into (1), we have: 82 + 02 + D(8) + E(0) + F = 0 i.e. 8D + F = –64 ……(2) By substituting (0, 6) into (1), we have: 02 + 62 + D(0) + E(6) + F = 0 i.e. 6E + F = –36 ……(3) By substituting (0, 0) into (1), we have: 02 + 02 + D(0) + E(0) + F = 0 i.e. F=0 By substituting F = 0 into (2), we have: 8D + 0 = –64 D = –8 By substituting F = 0 into (3), we have: 6E + 0 = –36 E = –6 ∴ The equation of the circle is x2 + y2 – 8x – 6y = 0.

(a) The equation of the circle is (x – 0)2 + (y – 28)2 = 202 x2 + y2 – 56y + 384 = 0 ……(1) (b) By substituting x = 12 into (1), we have: 122 + y2 – 56y + 384 = 0 y2 – 56y + 528 = 0 (y – 44)(y – 12) = 0 y = 44 or y = 12 (rejected) ∴ The position of the person is (12, 44). 35. (a) The x-axis is tangent to the circle. ∴ PC ⊥ x-axis PC // y-axis ∴ PC = 5 PB = PC (radii) ( a −0) 2 +(5 −2) 2 = 5 a2 + 9 = 25 a2 – 16 = 0

48

12 Coordinate Treatment of Simple Locus Problems

(a – 4)(a + 4) = 0 a= 4

or

a = –4 (rejected)

(b) The equation of the circle is (x – 4)2 + (y – 5)2 = 52 x2 + y2 – 8x – 10y + 16 = 0 (c) By substituting x = 0 into x2 + y2 – 8x – 10y + 16 = 0, we have: 02 + y2 – 8(0) – 10y + 16 = 0 y2 – 10y + 16 = 0 (y – 2)(y – 8) = 0 y=2 or y = 8 ∴ The coordinates of A =(0, 8)

( −2)   ( −4) − ,−  36. (a) Centre of C =  2 2   1 = ( 2,1) 2

2

 −4   −2  Radius of C =  2  +  2  −1     1 =2 4  ( −12 ) − ,−  Centre of C =  2 2  2 = (6,−2) 2

2

 −12  4 Radius of C =  2  +  2  − 31     2 =3

(d)

(b) Distance between the two centres =

=5 =2+3 = radius of C1 + radius of C2 ∴ The two circles touch each other.

Draw AP and BP. With the notations in the figure,

= sinθ

=

 AB     2 

Multiple Choice Questions (p. 94)

AP

1.

6    2 

5 3 = 5

Answer: D ∵ The two lines are parallel. ∴ They have the same slope.

−

2 1 =− ( −k ) 3 k = –6

θ = 36.87° ∠ APB = 2θ

∴

2.

= 2 × 36.87° = 73.74° Area of sector PAKB = π × 52 ×

73 .74 ° 360 °

= 16.09

1 × AP × BP × sin 73.74° 2 1 = × 5 × 5 × sin 73.74° 2

Area of PAB =

∴

( 2 −6) 2 +[1 −( −2)] 2

Answer: B Since the line y = 5 passes through A, the coordinates of A are (0, 5). Put x = 0 into x – 2y + 4 = 0, we have: 0 −2 y + 4 = 0 y =2 ∴ The coordinates of B are (0, 2). Put y = 5 into x – 2y + 4 = 0, we have:

x − 2(5) + 4 = 0

= 12.00 Area of segment AKB = area of sector PAKB – area of PAB = 16.09 – 12.00 = 4.1 (cor. to 1 d.p.)

x =6

∴

The coordinates of C are (6, 5).

∴

Area of ABC =

1 × AB × AC 2 1 = × (5 – 2) × (6 – 0) 2 =9

3.

Answer: C Let L be the line joining (–3, 2) and (4, –1).

49

Certificate Mathematics in Action Full Solutions 5A

−1 − 2 4 − ( −3) Slope of L 3 =− 7

y-intercept of L = −

=

Slope of L = −

The required line should have slope 4.

7 . 3

7.

Answer: C Put x = 2 into 3x + 5y = 26, we have: 3( 2) +5 y = 26 ∴

y =4 The coordinates of A are (2, 4).

5.

AC

= (7 −2) 2 +(1 −4) 2

Answer: B x + 2y + 8 = 0 ……(1) 3x + ay – 11 = 0 ……(2) By substituting (b, –5) into (1), we have: b + 2(–5) + 8 = 0 b=2 ∴ The point of intersection are (2, –5).

8.

Answer: A

= 34

Answer: D Slope of L1 = − Slope of L2 =

∴ ∴

4 =2 ( −2)

1 2

Slope of L3 = −

4 = –2 ≠ 2 2

∴ ∴

∵ ∴

L1 is not parallel to L3. B is false.

Slope of L1 × slope of L2 = 2 ×

 −4 + 6 10 + ( −4)  ,  = (1, 3) 2  2 

1 = 1 ≠ –1 2

Slope of BD = slope of BP =

L1 is not perpendicular to L2. A is false.

( −5) 5 = 4 4 ( −5) 5 = x-intercept of L3 = − 4 4

the x-axis.

Answer: C From the figure, x-intercept of L < 0.

−

c <0 1

c>0 y-intercept of L > 0 ∴

5 , 0) which is on 4

−

c >0 b

b<0 10. Answer: A

Answer: D x-intercept of L = −

9.

∴

x-intercept of L1 = x-intercept of L3 L1 and L3 intersect at the point (

3 −2 1 = 1 −( −8) 9

Only equation of option A has the same slope.

x-intercept of L1 = −

∴

Diagonals bisect each other. The mid-point P of A and C lies on BD. P=

Slope of L1 × slope of L3 = 2 × –2 = –4 ≠ –1 ∴ L1 is not perpendicular to L3. ∴ C is false.

6.

a <0 b

By substituting (2, –5) into (2), we have: 3(2) + a(–5) – 11 = 0 a = –1

Put y = 1 into 3x + 5y = 26, we have: 3x + 5(1) = 26 x=7 ∴ The coordinates of C are (7, 1). ∴

c <0 b

c <0 a

Centre of C

 k (−8)   k  =  − ,−  =  − ,4  2   2   2

Since L divides C into two equal parts, L passes through the centre of C.

50

12 Coordinate Treatment of Simple Locus Problems

 k  − , 4  into the equation of L, we  2 

By substituting

 

k  +3(4) – 5 = 0 2 k=7

11. Answer: D B and C do not represent equations of circle since the coefficients of x2 and y2 are not equal. 2

∴

2

 4  −6   +  − 20 = − 7 2  2 

For A, radius =

=4

(−1 −3 ) 2 +[ −1 −( −5)] 2 2

>5 = radius of circle C ∴ (–1, –1) lies outside the circle. ∴ III is true.

have: 2 −

=

A represents an imaginary circle. 2

1 2 1  − 2    2   =  + −    2  2   2  For D, radius 3 = >0 4

15. Answer: D By substituting (4, –1) into the equation of C, we have: L.H.S. = 42 + (–1)2 – 6(4) + 4(–1) + 5 = –6 ≠ 0 ∴ I is false.

 

Centre of C =  −

( −6) 4 ,−  2 2

= (3, –2) ∴ II is true. By substituting y = 0 into the equation of C, we have: x2 + 02 – 6x + 4(0) + 5 = 0 x2 – 6x + 5 = 0 (x – 1)(x – 5) = 0 x = 1 or x = 5 ∴ III is true.

∴ D represents a real circle. 12. Answer: C Since the circle touches the x-axis, radius = 2. The required equation is (x – 3)2 + [y – (–2)]2 = 22 x2 + y2 – 6x + 4y + 9 = 0 13. Answer: C Centre of C1 = centre of C2



= −



( −10 ) ( −4)  ,−  2 2 

∵

= (5, 2) C1 passes through the origin.

∴

Radius =

(5 −0) 2 +( 2 −0) 2 = 29

The equation of C1 is (x – 5)2 + (y – 2)2 = ( 29 ) 2 x2 + y2 – 10x – 4y = 0 14. Answer: C

 

Centre of C =  − ∴

( −6) 10  ,−  2 2 

= (3, –5) I is false. 2

Radius =

2

 −6   10    +  −9 2    2 

=5 ∴ II is true. The distance between (–1, –1) and the centre

51