5ach12(coordinate Treatment Of Simple Locus Problems)

12 Coordinate Treatment of Simple Locus Problems

12

Coordinate Treatment of Simple Locus Problems

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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

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.

(a)

y −5 x −3

(a) (b) ∵ ∴ 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.

y −5 =2 x −3 y = 2x – 1

y − y1 =m x − x1 y – y1 = m(x – x1)

Activity 12.3 (p. 56) 1.

(a)

(a) Ax + By + C = 0 By = –Ax – C A C y =− x− B B (b) Slope = −

2.

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

3.

A C , y-intercept = − B B

By substituting (p, 0) into the equation Ax + By + C = 0, we have: A(p) + B(0) + C = 0 C p=− A C ∴ x-intercept = − A 1 Slope = 2, y-intercept = 1, x-intercept = − 2

7

Certificate Mathematics in Action Full Solutions 5A

Activity 12.4 (p. 70)

The equation of L5 is y =

4.

The equation of L7 is y =

( x − 0) 2 + ( y − 0) 2 = 3

1.

∴ x2 + y2 = 32 ∴ The equation of the circle is x2 + y2 = 9. ( x − 1) 2 + ( y − 2) 2 = 3

2.

∴ (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

3.

1.

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

5.

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

6.

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

Activity 12.5 (p. 82)

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 y 2 − y1 x 2 − x1 By the point-slope form, y 2 − y1 y – y1 = (x – x1) x 2 − x1 Slope of L2 =

∴ 3.

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

p. 49

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

1.

1 x. 2 The equation of L6 is y = –x.

3.

y − y1 y − y1 = 2 x − x1 x 2 − x1

The straight line L3 cuts the x-axis at (a, 0), and the y-axis at (0, b). b−0 Slope of L3 = 0−a b =− a By the point-slope form, b y – 0 = − (x – a) a bx + ay = ab x y + =1 ∴ a b

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.

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

p. 53 1.

Let m be the slope of the straight line L. 7 −3 m= 5 −1 =1 The equation of the straight line L is y – 3 = 1(x – 1) ∴ y=x+2

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12 Coordinate Treatment of Simple Locus Problems

2.

3.

4.

5.

6.

Let m be the slope of the straight line L. 4−3 m= − 3 −1 1 =− 4 The equation of the straight line L is 1 y – 3 = − (x – 1) 4 1 13 ∴ y =− x+ 4 4 Let m be the slope of the straight line L. 0−5 m= 3−0 5 =− 3 The equation of the straight line L is 5 y – 5 = − (x – 0) 3 5 ∴ y=− x+5 3 Let m be the slope of the straight line L. 0 − (−2) m= 4−0 1 = 2 The equation of the straight line L is 1 y – (–2) = (x – 0) 2 1 ∴ y= x–2 2 Let m be the slope of the straight line. 8−6 m= 5 − ( −3) 1 = 4 The equation of the straight line is 1 y – 8 = (x – 5) 4 1 27 ∴ y= x+ 4 4 Let m be the slope of the straight line. −5 − 0 m= 0 − (−8) 5 =− 8 The equation of the straight line is 5 y – (–5) = − (x – 0) 8

∴

y=−

5 x–5 8

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)

2.

y+2 2 =− x+3 3 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 ∴ Slope = − B 2 =− 1 = −2 C A 0 =− 2 =0

x-intercept = −

C B 0 =− 1 =0

y-intercept = −

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 ∴ Slope = −2 y-intercept = 0 (b) From the equation 4x – 5y + 15 = 0, we have A = 4, B = –5 and C = 15. A ∴ Slope = − B

9

Certificate Mathematics in Action Full Solutions 5A

 4   = −  −5 4 = 5 C A 15 =− 4

x-intercept = −

C B  15   = −  −5 =3

y-intercept = −

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: 4 y= x+3 5 4 ∴ Slope = 5 y-intercept = 3 (c) From the equation 3x + 5y + 5 = 0, we have A = 3, B = 5 and C = 5. A ∴ Slope = − B 3 =− 5 C A 5 =− 3

x-intercept = −

C B 5 =− 5 = −1

y-intercept = −

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

3 x + 5(0) + 5 = 0 3x = −5 5 x=− 3 5 ∴ x-intercept = − 3 If we make y the subject of the equation, we have: 3 y=− x–1 5 3 ∴ Slope = − 5 y-intercept = −1 (d) From the equation 6x – 2y – 7 = 0, we have A = 6, = –2 and C = –7. A ∴ Slope = − B  6   = − −2 =3

B

C A  −7  = −   6  7 = 6

x-intercept = −

C B −7 = −   − 2 7 =− 2

y-intercept = −

Alternative Solution Put y = 0 into 6x – 2y – 7 = 0, we have: 6 x − 2(0) − 7 = 0 6x = 7 7 x= 6 7 ∴ x-intercept = 6 If we make y the subject of the equation, we have: 7 y = 3x – 2 3 ∴ Slope = y-intercept = −

7 2

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12 Coordinate Treatment of Simple Locus Problems

p. 62 1.

(ii) Put y = 0 into y =

From the equation of L: 3x – 4y + 10 = 0, we have: 3 3 = Slope of L = − (−4) 4 (a) ∵ L1 // L ∴ Slope of L1 = slope of L 3 = 4 ∴ The equation of L1 is 3 y – 3 = (x – 1) 4 3 9 y = x+ 4 4

5 x+3 2 6 x=− 5 6 ∴ x-intercept = − 5 0=

p. 67 1.

(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 y – 3 = − (x – 0) 3 4 y=− x+3 3 2.

(b) (i) ∵ L2 ⊥ L ∴ Slope of L2 × slope of L = –1  2 Slope of L2 ×  −  = –1  5 5 Slope of L2 = 2 ∴ The equation of L2 is 5 y – 3 = (x – 0) 2 5 y= x+3 2

(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 1 y= 5 1 ∴ The coordinates of A = (0, ) 5 1 5 =4 1− 0 5 ∴ The equation of L is 4 y – 1 = (x – 1) 5 4 1 y = x+ 5 5

(b) Slope of L =

From the equation of L: 2x + 5y – 7 = 0, we have: 2 Slope of L = − 5 (a) ∵ L1 // L ∴ Slope of L1 = slope of L 2 =− 5 ∴ The equation of L1 is 2 y – 0 = − [x – (–2)] 5 2 4 y =− x− 5 5

5 x + 3, we have: 2

2.

1−

(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) 1 − ( −2) 3 =− 1− 3 2 ∴ The equation of L is 3 y – 1 = − (x – 1) 2 3 5 y =− x+ 2 2

(b) Slope of L =

3.

(a) L1: 2x – 3y – 4 = 0 L2: 3x + y + 5 = 0

……(1) ……(2)

11

Certificate Mathematics in Action Full Solutions 5A

(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 = ( −1, − 2)

3.

4 3 3

(e)

centre: (0, 0), radius =

(f)

centre: (–4, 0), radius = 3 2

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 1 − ( −2) 3 = 1 − ( −1) 2 ∴ The equation of L is 3 y – 1 = (x – 1) 2 3 1 y = x− 2 2

∴ Matching is as follows:

(b) Slope of L =

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 = (1, 4) (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. 75 1.

2.

3.

4.

5.

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

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

(c)

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

(d) (x + 2)2 + (y – 5)2 = 8 (e) (x + 4)2 + (y + 5)2 =

(x – 1)2 + (y – 2)2 = 32 x + y2 – 2x – 4y – 4 = 0 2

2

(c) x + (y + 3) = 16

(a)

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

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

(b) (x – 4)2 + (y – 3)2 = 4

2.

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

x 2 + y 2 + 2 2 x + 2 3 y − 12 = 0

(a) x2 + y2 = 7

2

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

p. 72 1.

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

9 4

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

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

(b)

centre: (2, 3), radius = 6

(c)

centre: (0, 1), radius = 5

(d)

centre: (–1, –5), radius = 10

6.

(a) Centre

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

2

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

12

12 Coordinate Treatment of Simple Locus Problems

(b) Centre

∴ The equation of the circle is

 8 0 =  − ,−   2 2 = ( −4,0) 2

(x – 3)2 + [y – (–5)]2 = ( 5 ) 2 x2 + y2 – 6x + 10y + 29 = 0 2

8 0 Radius =  2  +  2  − (−3) = 19

(c) Centre

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

2

 0   − 12  Radius =  2  +  2  − 20     =4 (d) 2x2 + 2y2 – 4x – 7 = 0 7 x2 + y2 – 2x – = 0 2 ( − 2 ) 0  = − ,−  Centre  2 2 = (1,0) 2

2

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

3 2 2

(e) 4x2 + 4y2 + 12x – 20y + 8 = 0 x2 + y2 + 3x – 5y + 2 = 0  3 ( −5)  =  − ,−  2  Centre  2  3 5 = − ,   2 2 2

Radius

2

 3   −5 =   +  −2 2  2  =

26 2

p. 79 1.

(a) The coordinates of

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

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

2.

(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 59 D= 19 59 By substituting D = into (5), we have: 19  59  3  – E = 8  19  25 E= 19 25 By substituting E = into (4), we have: 19  25  –4   + F = –16  19  204 F=− 19 ∴ The equation of the circle is 59 25 204 x2 + y2 + x+ y− = 0. 19 19 19 (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

13

Certificate Mathematics in Action Full Solutions 5A

0−3 3 − (−1) 3 =− 4

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.

Slope of AB =

Let θ be the inclination of AB. ∵ Slope of AB = tan θ 3 − = tan θ 4 ∴ θ = 180° − 36.87° = 143.1° (cor. to 1 d.p.)

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 2x + 3y – 12 = 0

2.

∴ The inclination of AB is 143.1°. PQ = ( −2 − 4) 2 + ( −1 − 2) 2 (b)

(a) The equation of the straight line is y − ( −3) −4 − (−3) = x−2 −1 − 2 y + 3 −1 = x −2 −3 3y + 9 = x − 2 x − 3 y − 11 = 0

=3 5 −1 − 2 −2−4 1 = 2

Slope of PQ =

Let θ be the inclination of PQ. ∵ Slope of PQ = tan θ 1 = tan θ ∴ 2 θ = 26.6° (cor. to 1 d.p.)

(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 x+ y+9 = 0 3.

∴ The inclination of PQ is 26.6°. MN = [−3 − ( −1)]2 + [2 − ( −4)]2 (c)

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

2 − (−4) − 3 − ( −1) = −3

Slope of MN =

Let θ be the inclination of MN. ∵ Slope of MN = tan θ −3 = tan θ ∴ θ = 180° − 71.57° = 108.4° (cor. to 1 d.p.) ∴ The inclination of MN is 108.4°.

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

= ( −2) 2 + 6 2 = 2 10

(b) The equation of the straight line is x y + =1 −2 1   2 x − 4y + 2 = 0

1.

= ( −6) 2 + (−3) 2

2.

Perimeter of ABC = AB + BC + CA = ( 26 − 1) 2 + (1 − 1) 2 + (10 − 26) 2 + (13 − 1) 2 +

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

(1 − 10) 2 + (1 − 13) 2 = 25 + 20 + 15 = 60

14

12 Coordinate Treatment of Simple Locus Problems

3.

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

7.

( x − 5) 2 + (0 − 4) 2 = 5 (x – 5)2 = 9 x=8

or x = 2 ( 8 , 0 ) ∴ The coordinates of B = or ( 2, 0) 4.

∵ P, Q and R are collinear. ∴ Slope of PQ = slope of QR −1 − 4 −8 − (−1) = 3−2 x −3 5 x − 15 = 7 22 x= 5

5.

(a) ∵ AB // CD ∴ Slope of AB = slope of CD 7 − 2 y − (−1) = 4−0 7−2 25 = 4 y + 4 21 y= 4

(b) Let (0, c) be the coordinates of C. ∵ CM ⊥ AB ∴ Slope of CM × slope of AB = –1 c − 4 7 −1 × = −1 0−2 4−0 4 c−4= 3 16 c= 3 16 ∴ The coordinates of C = (0, ) 3 8.

(b) ∵ AC ⊥ BD ∴ Slope of AC × slope of BD = –1 −1 − 2 y − 7 × = −1 2−0 7−4 y −7 = 2 y=9 6.

(a) Let (x, y) be the coordinates of B. ∵ M is the mid-point of AB. 2+ x 4+ y ∴ =5 and =2 2 2 x=8 and y=0 ( 8 , 0 ) ∴ The coordinates of B = 4−0 2−0 =2 ∵ CM // OA ∴ Slope of CM = slope of OA =2

(b) (i) Slope of OA =

(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

(ii) Let (0, y) be the coordinates of C. ∵ Slope of CM = 2 2− y =2 ∴ 5−0 y = −8 ∴ The coordinates of C = (0, − 8)

∴ The coordinates of C = (0, 12) 9. (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°.

(a) Let (x, y) be the coordinates of M. ∵ M is the mid-point of AB. 0+4 1+ 7 y= ∴ x= and 2 2 x=2 and y=4 ∴ The coordinates of M = ( 2, 4)

(a) Let (x, y) be the coordinates of P. By the section formula for internal division, we have: 1(4) + 2(7) x= 2 +1 =6 1(6) + 2(9) y= 2 +1 =8 ∴ The coordinates of P = (6, 8)

15

Certificate Mathematics in Action Full Solutions 5A

(b) Let (x, y) be the coordinates of P. By the section formula for internal division, we have: 4( −5) + 3(2) x= 3+ 4 = −2 4(6) + 3( −2) y= 3+ 4 18 = 7 18 ∴ The coordinates of P = (−2, ) 7 (c) Let (x, y) be the coordinates of P. By the section formula for internal division, we have: 3( −13) + 7(7) x= 7 + 13 =1 3(0) + 7(10) y= 7+3 =7 ∴ The coordinates of P = (1, 7) 10. (a) Let AP : PB = r : s. By the section formula for internal division, we have: s ( −1) + r (−6) −4= r+s − 4r − 4 s = − s − 6 r 2r = 3s r 3 = s 2 ∴ AP : PB = 3 : 2 (b) By the section formula for internal division, we have: 2( −2) + 3(7) y= 3+ 2 17 = 5 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. x +x y + y3 x + x4 y + y4 ,2= 2 ∴ 1= 1 3 , 2 = 1 and 1 = 2 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).

Level 2 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 25 ∴ The coordinates of C = (0, ) 4 (b) Slope of AB × slope of OA 10 − 6 6 = × 5 −8 8 4 3 = × −3 4 = −1 ∴ AB ⊥ OA 2 2 (c) OA = (8 − 0) + (6 − 0) = 10

AB = (5 − 8) 2 + (10 − 6) 2 =5 25   CB = (5 − 0) 2 + 10 −  4  25 = 4

2

Area of trapezium OABC 1 = × (CB + OA) × AB 2 1  25  = ×  + 10  × 5 2  4  = 40.625 14. (a) (i) ∵ AB ⊥ BC ∴ Slope of AB × slope of BC = –1 5−4 c −5 × = −1 3 − (−1) 5 − 3 c − 5 = −8 c = −3

16

12 Coordinate Treatment of Simple Locus Problems

(ii) Area of ABC 1 = × AB × BC 2 1 = × [3 − (−1)]2 + (5 − 4) 2 × (5 − 3) 2 + ( −3 − 5) 2 2 1 = × 17 × 2 17 2 = 17 (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 17 x=− 6 17 ∴ The coordinates of D = (− , 0) 6

y − ( −4) = (8 − 0) + (0 − y )

 −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

15. (a) Let (0, y) be the coordinates of A. ∵ AB = AC 2

 −2 + (−3) 6 + 3  P= ,  2 2   16. (a) The coordinates of  5 9 = − ,   2 2

(b) (i) 2

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

∴ The coordinates of A = (0, 6)

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

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

(c) Area of ABC 1 = × BC × AM 2 1 = × (8 − 0) 2 + [0 − ( −4)]2 × ( 4 − 0) 2 + (−2 − 6) 2 2 1 = × 80 × 80 2 = 40

2

= 22 + 22 = 2 2 2

 0 + 8 −4 + 0  (b) The coordinates of M =  2 , 2  = (4,−2) Slope of AM × slope of BC −2 − 6 0 − (−4) = × 4−0 8−0 −8 4 = × 4 8 = −1 ∴ AM ⊥ BC

2

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

2

= 32 + (−3) 2 =3 2 2

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

2

= 32 + ( −3) 2 =3 2 ∴ PS = QR and PQ = SR (ii) Slope of PS × slope of PQ 13 9 3 9 − − 2 2 = × 2 2 1  5 1  5 − −−  −−  2  2 2  2  2 −3 = × 2 3 = −1 ∴ PS ⊥ PQ

17

Certificate Mathematics in Action Full Solutions 5A

Slope of QR × slope of SR 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 14 ∴ The coordinates of D = (0, ) 5

∴

∴ The coordinates of E = ( 4,

22 . 5

22 ) 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 ∴ The coordinates of H = (5, 2) (b) ∵ PH ⊥ QR ∴ Slope of PH × slope of QR = –1 2 − 7 3 − (−2) × = −1 5 − a 7 − (−3) 5 = 10 − 2a 5 a= 2 2

5  PH =  5 −  + (2 − 7) 2 2  =

2−6 4 =− 7−2 5

5 5 2

(c) Area of PQR 1 = × QR × PH 2

∵ DE // BC ∴ Slope of DE = slope of BC 2 − (−2) = 7 − (−3) 2 = 5 (ii) Let (x, y) be the coordinates of E. ∵ Slope of AE = slope of AC y −6 4 =− ∴ x−2 5 5y – 30 = –4x + 8 4x + 5y – 38 = 0 ……(1) 2 ∵ Slope of DE = 5

……(2)

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

(ii) Let AD : DB = r : s. By the section formula for internal division, we have: s ( 2) + r (−3) 0= r+s 2 s = 3r r 2 = s 3 ∴ AD : DB = 2 : 3 (b) (i) Slope of AC =

14 5 =2 x−0 5 2x – 5y + 14 = 0 y−

=

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

1 5 5 ×5 5 × 2 2 125 = 4 =

Exercise 12B (p. 44) Level 1 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.

18

12 Coordinate Treatment of Simple Locus Problems

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

6.

By substituting (–4, c) into the equation y = 2, we have: c=2

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

8.

∵ Slope = −

9.

i.e. y = –3x (d) The equation of the straight line is −9 y= x 12 3 i.e. y = − x 4

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

7.

1 3

The equation of L is y = −

4.5 x − 1.5

Level 2

The equation of L is 2 y= x −4 1 y=− x 2 i.e.

∴

y=

1 x. 3

(a) The equation of the straight line is y = 6. (b) The equation of the straight line is y = 0. (c) The equation of the straight line is y = –3. (d) The equation of the straight line is y = 7.

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

13. ∵ L passes through A and B. 3 By substituting (a, 6) into the equation y = − x , we 4 have: 3 6=− a 4 ∴ a = −8 3 By substituting (6, b) into the equation y = − x , we 4 have: 3 b = − (6) 4 9 ∴ b=− 2 14. (a)

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

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

(d) The equation of the straight line is x = 0. 11. (a) The equation of the straight line is 8 y= x 2 i.e. y = 4x (b) The equation of the straight line is −5 y= x −4 5 i.e. y = x 4 (c) The equation of the straight line is

(b) (i) By joining the points in (a), we can see that L passes through the origin. y − coordinates ∴ is a constant. x − coordinates y 3 = ∴ x 2 y : x = 3: 2

19

Certificate Mathematics in Action Full Solutions 5A

Level 1 (ii)

y 3 From (b) (i), we have = x 2 3 y= x 2 ∴

15.

1. 2.

3 The equation of L is y = x . 2

(a) Let (0, y), (h, k) and (x, 0) be the coordinates of A,

3.

The equation of the straight line L is 1 y – 4 = [x – (–2)] 2 1 ∴ y= x+5 2

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 1 y = − x + (–3) 2 1 ∴ y=− x–3 2

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 4 y – 5 = − [x – (–2)] 5 4 17 ∴ y =− x+ 5 5

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 ∵ 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 y= x, −3 y = −x i.e. 16. (a) The equation of L1 is −2 y= x, −3 i.e. 2 y= x 3 The equation of L2 is −3 y= x, 1 i.e. y = −3 x (b) Let θ1 and θ2 be the inclinations of L1 and L2 respectively. tan θ1 = slope of L1 2 3 θ1 = 33.7° =

tan θ 2 = slope of L2 = −3 θ 2 = 180° − 71.6° = 108.4° θ2 – θ1 = 108.4° – 33.7° = 75° (cor. to the nearest degree) ∴ The acute angle between L1 and L2 is 75°.

Exercise 12C (p. 50)

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

10. The equation of the straight line is   1  y – 3 = 0 x −  −    3  ∴

y=3

11. The equation of the straight line is y = 3x + 4 12. The equation of the straight line is 2 y=− x+1 5 13. The equation of the straight line is 4 4 y = x+ 7 7

20

12 Coordinate Treatment of Simple Locus Problems

y-intercept = 3

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 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).

(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 (c) ∵ Slope of L3 = =

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

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.

3 × slope of L1 4 3 = ×2 4 3 = 2 y-intercept = c + 2 = 4 + 2 = 6

(b) (i) Slope of L2 =

∴ The equation of L2 is y =

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

1 x+3 , 2

3 x + 6. 2

(ii) By substituting (2a, a) into the equation 3 y = x + 6 , we have: 2 3 a = (2a) + 6 2 − a= 3

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

(ii) From the equation of L1: y = −

1  1 −   2

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

Level 2

19. (a) (i) The equation of the straight line L1 is 1 y – 4 = − [x – (–2)] 2 1 ∴ y=− x+3 2

1 slope of L1

Exercise 12D (p. 53) Level 1 1.

Let m be the slope of the straight line L. 1− 3 m= 2 −1 = −2 The equation of the straight line L is

21

Certificate Mathematics in Action Full Solutions 5A

∴

2.

3.

4.

5.

6.

y – 1 = –2(x – 2) y = –2x + 5

Let m be the slope of the straight line L. 1 − (−1) m= 3 − ( −2) 2 = 5 The equation of the straight line L is 2 y – 1 = (x – 3) 5 2 1 ∴ y = x− 5 5 Let m be the slope of the straight line L. −7 − ( −8) m= 6 − ( −7) 1 = 13 The equation of the straight line L is 1 y – (–7) = (x – 6) 13 1 97 ∴ y = x− 13 13 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 y – 3 = − (x – 0) 2 1 ∴ y=− x+3 2 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 y – 2 = (x – 0) 5 2 ∴ y= x+2 5 Let m be the slope of the straight line L.

0 − (−3) −5−0 3 =− 5 The equation of the straight line L is 3 y – (–3) = − (x – 0) 5 3 ∴ y=− x–3 5 m=

7.

Let m be the slope of the straight line. 0 − ( −2) m= 3 − (−1) 1 = 2 The equation of the straight line is 1 y – 0 = (x – 3) 2 1 3 ∴ y = x− 2 2

8.

Let m be the slope of the straight line. −1 − ( −3) m= 2 −1 =2 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) m= − 3 − (−4) = −1 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) m= 1 − (−5) 4 = 3 The equation of the straight line is 4 y – 1 = (x – 1) 3 4 1 ∴ y = x− 3 3 11. Let m be the slope of the straight line. 1− 0 m= 0−3 1 =− 3

22

12 Coordinate Treatment of Simple Locus Problems

The equation of the straight line is 1 y – 1 = − (x – 0) 3 1 ∴ y=− x+1 3 12. Let m be the slope of the straight line. 5 −0 m= 2 0−5 1 =− 2 The equation of the straight line is 1 5 y – = − (x – 0) 2 2 1 5 ∴ y =− x+ 2 2 13. Let m be the slope of the straight line. 5−0 m= 0 − (−4) 5 = 4 The equation of the straight line is 5 y − 5 = ( x − 0) 4 5 y = x+5 ∴ 4 14. Let m be the slope of the straight line. −4 − 0 m= 0 − ( −3) 4 =− 3 The equation of the straight line is 4 y – (–4) = − (x – 0) 3 4 ∴ y=− x–4 3

Level 2 15. (a)

Let m be the slope of the straight line L. 6 − ( −2) m= −5−5 4 =− 5 The equation of the straight line L is 4 y – (–2) = − (x – 5) 5 4 ∴ y=− x+2 5

(b) From the equation of L: y = −

4 x + 2, 5

y-intercept = 2

16. (a)

Let m be the slope of the straight line L. −2 − 0 m= 0−a 2 = a The equation of the straight line L is 2 y = x + (–2) a 2 ∴ y= x–2 a 1 3 2 1 = a 3 a =6

(b) ∵ Slope of L = ∴

17. (a) The equation of the straight line L1 is 4 y – (–3) = [x – (–2)] 3 4 1 y = x− 3 3 The equation of the straight line L2 is 4 y – 2 = (x – 4) 3 4 10 ∴ y = x− 3 3 (b) (i) Let m be the slope of the straight line. 2 − ( −3) m= 4 − ( −2) 5 = 6 The equation of the required straight line is 5 y – 2 = (x – 4) 6 5 4 ∴ y = x− 6 3 (ii) From the equation of the required straight line: 5 4 y = x− , 6 3 4 y-intercept = − 3

23

Certificate Mathematics in Action Full Solutions 5A

8 x + 3(0) + 24 = 0 8 x = −24 x = −3 ∴ x-intercept = −3 If we make y the subject of the equation, we have: 8 y=− x–8 3 8 ∴ Slope = − 3

18. Let the x-intercept be 2a, then the y-intercept is a. Let m be the slope of the straight line. a −0 m= 0 − 2a 1 =− 2 The equation of the straight line is 1 y – 2 = − (x – 1) 2 1 5 ∴ y =− x+ 2 2

y-intercept = −8 6.

Exercise 12E (p. 59) Level 1 1.

4y = –3x + 2 ∴ 3x + 4y – 2 = 0

2.

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

3.

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

4.

5.

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

x-intercept = −

y−4 1 =− x−2 2 2y – 8 = –x + 2 ∴ x + 2y – 10 = 0 From the equation 8x + 3y + 24 = 0, we have A = 8, 3 and C = 24. A =− B ∴ Slope 8 =− 3 C A x-intercept 24 =− 8 = −3

=7

Alternative Solution If we make y the subject of the equation, we have: 7 y=− x+7 2 7 ∴ Slope = − 2

B=

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

=−

C B y-intercept 24 =− 3 = −8 =−

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

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

7.

From the equation 2x – 3y + 9 = 0, we have A = 2, B = –3 and C = 9. A =− B ∴ Slope = − 2   −3 2 = 3

24

12 Coordinate Treatment of Simple Locus Problems

x-intercept

C A 9 =− 2 =−

∴ Slope = −

y-intercept =

C B y-intercept  9  = −   −3 =3

Alternative Solution

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 9 ∴ x-intercept = − 2 8.

From the equation 3x + 7y – 2 = 0, we have A = 3, B = 7 and C = –2. A =− B ∴ Slope 3 =− 7 C =− A  x-intercept = − − 2   3  2 = 3 C B y-intercept = − − 2   7  2 = 7 =−

Alternative Solution If we make y the subject of the equation, we have: 3 2 y =− x+ 7 7

2 7

Put y = 0 into 3x + 7y – 2 = 0, we have: 3 x + 7(0) − 2 = 0 3x = 2 2 x= 3 2 ∴ x-intercept = 3

=−

If we make y the subject of the equation, we have: 2 y= x+3 3 2 ∴ Slope = 3

3 7

9.

A 3 3 A − =− 4 3 9 = 4A 9 A= 4

∵ Slope = −

∴

(−6) B 2 ( −6) =− 3 B 2 B = 18 B=9

10. ∵ y-intercept = −

∴

C 2 C −4= − 2 C =8

11. ∵ x-intercept = − ∴

Level 2 12.

2(y + 3) = 3(x – 2) 2y + 6 = 3x – 6 3x – 2y – 12 = 0 A =− B ∴ Slope = − 3   −2 3 = 2 C A x-intercept  − 12  = −   3  =4 =−

25

Certificate Mathematics in Action Full Solutions 5A

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

13.

y x + =1 2 3 3y + 2x = 6 2x + 3y – 6 = 0 A =− B ∴ Slope 2 =− 3 C A x-intercept  −6 = −   2  =3 =−

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

14. 2(x – 3y) = x – 4y + 2 2x – 6y = x – 4y + 2 x – 2y – 2 = 0 A =− B ∴ Slope = − 1  −2 1 = 2 C A x-intercept −2 = −   1  =2 =−

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

15.

2y − 2 2 =− 3x + 1 3 6y – 6 = –6x – 2 6x + 6y – 4 = 0 3x + 3y – 2 = 0

A B ∴ Slope 3 =− 3 = −1 =−

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

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

16. For the straight line L1: x + 3y + C = 0, C y-intercept = − 3 For the straight line L2: 3x – 2y + 2 = 0,  2   =1 y-intercept = −  −2 C ∴ − 3 =1 C = −3 17. For the straight line L1: 2x + 5y + 6 = 0, 2 slope = − 5 For the straight line L2: x – 4y + 8 = 0,  8  y-intercept = − =2 −4 ∴ The equation of L is y = −

2 x + 2. 5

18. For the straight line L1: kx + 2y + 6 = 0, k slope = − 2 For the straight line L2: x + (k – 1)y + 3 = 0, 1 slope = − k −1 k 1 − =− ∴ 2 k −1 k(k – 1) = 2 k2 – k – 2 = 0 (k + 1)(k – 2) = 0 k = −1 or k = 2

26

12 Coordinate Treatment of Simple Locus Problems

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

1.

From the equation of L: x + y = 0, we have: 1 Slope of L = − 1 = −1 ∵ 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

2.

From the equation of L: x – 2y = 0, we have: 1 L=− ( −2) Slope of 1 = 2 ∵ The required straight line // L. 1 ∴ Slope of the required straight line = 2 ∴ The equation of the required straight line is 1 y – (–3) = [x – (–2)] 2 1 y= x–2 2

3.

From the equation of L: 2x – y + 3 = 0, we have: 2 Slope of L = − ( −1) =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

4.

From the equation of L: 2x + 3y – 4 = 0, we have: 2 Slope of L = − 3 ∵ The required straight line // L. 2 ∴ Slope of the required straight line = − 3 ∴ The equation of the required straight line is 2 y – 0 = − [x – (–5)] 3 2 10 y=− x− 3 3

5.

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.

6.

Let m be the slope of the required straight line.

a slope = − b

y-intercept = −

6 b

a =3 b a = –3b ……(1) 6 and − = 4 b 3 b=− 2 ∴

−

By substituting b = −

3 into (1), we have: 2

 3 a = −3 −   2 9 = 2 20. (a) Let (0, a) and (b, 0) be the coordinates of A and B respectively. 15 ∵ x-intercept = − 3 ∴ b = –5  15   ∵ y-intercept = −  −5 ∴ a=3 ∴ The coordinates of A and B are (0, 3) and (–5, 0) respectively. (b) y-intercept of L2 = y-intercept of L1 = 3 2 ∴ The equation of L2 is y = − x + 3. 3 (c) Put y = 0 into y = − 0=− x=

2 x + 3, we have: 3

2 x+3 3

9 2

∴ The coordinates of C = (

9 , 0). 2

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

∴ Area of △ABC =

Exercise 12F (p. 63) Level 1

27

Certificate Mathematics in Action Full Solutions 5A

From the equation of L: 2x + 3y = 0, we have: 2 Slope of L = − 3 ∵ The required straight line ⊥ L. ∴ m × slope of L = –1  2 m ×  −  = −1  3 3 m= 2 ∴ The equation of the required straight line is 3 y – (–2) = (x – 1) 2 3 7 y= x− 2 2 7.

8.

9.

Let m be the slope of the required straight line. From the equation of L: x + y – 5 = 0, we have: 1 Slope of L = − 1 = −1 ∵ 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: 2 Slope of L = − (−1) =2 ∵ The required straight line ⊥ L. ∴ m × slope of L = –1 m × 2 = –1 1 m=− 2 ∴ The equation of the required straight line is 1 y − 0 = − [ x − ( −2)] 2 1 y = − x −1 2 Let m be the slope of the required straight line. From the equation of L: 5x – 4y – 6 = 0, we have: 5 L=− ( −4) Slope of 5 = 4 ∵ The required straight line ⊥ L. ∴ m × slope of L = –1

5 = –1 4 4 m=− 5 ∴ The equation of the required straight line is 4 y – (–4) = − [x – (–2)] 5 4 28 y =− x− 5 5 m×

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 Slope of L1 = − (−1) = 2 ∵ 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: 2 Slope of L1 = − 3 ∵ L ⊥ L1 ∴ m × slope of L1 = –1  2 m ×  −  = −1  3 3 m= 2 ∴ 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 ∵ L ⊥ L1 ∴ m × slope of L1 = –1 4 m × = –1 3 3 m=− 4 ∴ The equation of L is

28

12 Coordinate Treatment of Simple Locus Problems

3 (x – 2) 4 3 11 y =− x+ 4 2

∴ The possible equation of BC is y = −

y–4=−

14. ∵ L1 // L2 ∴ Slope of L1 = slope of L2 a a+2 − =− 3 ( −2) 2a = −3a − 6 6 a=− 5

1 x + 12 (or any other reasonable 2 answers). (b) ∵ 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). y=−

(c) ∵ AD // BC ∴ Slope of AD = slope of BC = −

15. ∵ L1 ⊥ L2 ∴ Slope of L1 × slope of L2 = –1 a  4 − ×  −  = −1 ( − 2)  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 1 n=− m ∵ L1 intersects L2 at the y-axis. ∴ L1 and L2 have the same y-intercept. ∴ c=d 1 ∴ L1 : y = mx + c and L2 : y = − 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 1 y = − x − 4 or y = 2 x − 6 and y = − x − 6 2 respectively (or any other reasonable answers). 17. For AB: y = 2x – 8, slope = 2 Put y = 0 into y = 2x – 8, we have: 0 = 2x – 8 x=4 ∴ x-intercept = 4 (a) ∵ BC ⊥ AB ∴ Slope of BC × slope of AB = –1 Slope of BC × 2 = –1 1 Slope of BC = − 2

1 x + 10 or 2

1 2

∵ AD cuts the x-axis at A. ∴ x-intercept of AD = x-intercept of AB = 4 ∴ The equation of AD is 1 y – 0 = − (x – 4) 2 1 y=− x+2 2

Level 2 18. For L1: 2x + y + 3 = 0, 2 Slope = − = –2 1 3 x-intercept = − 2 ∴ x-intercept of L2 = −

3 2

∵ L1 ⊥ L2 ∴ Slope of L1 × slope of L2 = –1 –2 × slope of L2 = –1 1 slope of L2 = 2 ∴ The equation of L2 is 1  3  y − 0 =  x −  −  2  2  y=

1 3 x+ 2 4

19. For L1: 3x + y – 5 = 0, 3 Slope = − = –3 1 For L2: x + ay – 6 = 0, 1 Slope = − a For L3: bx + 3y – 28 = 0, b Slope = − 3 ∵ L1 ⊥ L2 ∴ Slope of L1 × slope of L2 = –1

29

Certificate Mathematics in Action Full Solutions 5A

 1 − 3 ×  −  = −1  a a = −3

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

∵ L1 // L3 ∴ Slope of L1 = slope of L3 b –3 = − 3 b =9

4 − (−2) = –1 −1 − 5 ∵ 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

(c) Slope of AC =

20. For L1: 4x + 3y – 6 = 0, 4 Slope = − 3 For L2: 2x – 3y – 5 = 0, 5  −5 y-intercept = − =− 3  −3 ∵ L // L1 L = slope of L1 ∴ Slope of 4 =− 3 5 y-intercept of L = − 3 ∴ The equation of L is y = −

23. (a) ∵ OR ⊥ OP ∴ Slope of OR × slope of OP = –1 3m = –1 1 m=− 3 4 5 x− . 3 3

21. For the straight line passing through B(–4, –1) and C(2, 3), 3 − ( −1) 2 = Slope = 2 − (−4) 3 ∵ L is perpendicular to the above straight line. 2 ∴ Slope of L × = –1 3 3 Slope of L = − 2 ∴ The equation of L is 3 y – 4 = − [x – (–2)] 2 3 y=− x+1 2 −2 − 0 1 =− 22. (a) Slope of BC = 5 −1 2 ∵ AD // BC ∴ Slope of AD = slope of BC = − ∴ The required equation is 1 y – 4 = − [x – (–1)] 2 1 7 y =− x+ 2 2 (b) Slope of AB =

4−0 = –2 −1−1

1 2

(b) (i) ∵ PQ // OR ∴ Slope of PQ = slope of OR = 3 ∴ The equation of PQ is y – 2 = 3(x – 8) y = 3x – 22 (ii) ∵ RQ // OP ∴ Slope of RQ = slope of OP = −

1 3

∴ The equation of RQ is 1 y − 2 = − ( x − 8) 3 1 14 y=− x+ 3 3 24. (a) Slope of OP =

3−0 3 =− −2−0 2

∵ OP ⊥ AB ∴ Slope of OP × slope of AB = –1 3 − × slope of AB = –1 2 2 Slope of AB = 3 The equation of L is 2 y – 3 = [x – (–2)] 3 2 13 y= x+ 3 3 (b) Rewrite the equation L: y =

2 13 x+ into the general 3 3

30

12 Coordinate Treatment of Simple Locus Problems

( x − y − 1) + ( x + y − 3) = 0 2x − 4 = 0 x=2 By substituting x = 2 into (1), we have: 2–y–1=0 y=1 ∴ The coordinates of A = (2,1)

form, we have 2x – 3y + 13 = 0. 13 x-intercept = − 2 1   13  39 × 0 −−  ×3 = ∴ Area of △OAP = 2   2  4 25. (a) For the equation L1: x + 2y – 8 = 0, (−8) x-intercept = − =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) (b) K =  2   2 ∵ 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

4.

L1: 3x – 2y = 6 ……(1) L2: y = –x ……(2) By substituting (2) into (1), we have: 3x – 2(–x) = 6 6 x= 5 6 By substituting x = into (2), we have: 5 6 y=− 5 6 6 ∴ The coordinates of A =  5 ,− 5   

5.

(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)

(c) Rewrite the equation L2: y = 2x – 6 into the general form, we have 2x – y – 6 = 0. (−6) x-intercept = − =3 2 1 ∴ Area of △AKC = × (8 – 3) × 2 2 =5 6.

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

L1: x = –3 L2: y = 2

∴ The coordinates of A = (−3, 2) 2.

3.

L1: 2x – 3y – 6 = 0 ……(1) L2: y = –3 ……(2) By substituting (2) into (1), we have: 2x – 3(–3) – 6 = 0 3 x=− 2  3  ∴ The coordinates of A =  − 2 , − 3    L1: x – y – 1 = 0 L2: x + y – 3 = 0 (1) + (2),

(b)

……(1) ……(2)

The equation of L is y =

3 x. 7

(a) L1: 3x + y – 3 = 0 ……(1) L2: 4y + 3 = 0 ……(2) From (2), we have: 3 y=− 4 3 By substituting y = − into (1), we have: 4  3 3x +  −  – 3 = 0  4 5 x= 4 5 3 ∴ The coordinates of A =  4 , − 4   

(b) Slope of L =

−

3 −0 4 5 −0 4

=−

3 5

The equation of L is y = −

3 x. 5

31

Certificate Mathematics in Action Full Solutions 5A

7.

(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 ∴ The coordinates of A = (−2, 3) (b) The equation of L is y = −

8.

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: –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 L2: y = 2x – 3 ……(2) By substituting (2) into (1), we have: 2 2x – 3 = x – 2 3 3 x= 4 3 By substituting x = into (2), we have: 4

10. Slope of L1 =

3 y = 2  – 3 4 3 =− 2 3 3 ∴ The coordinates of A =  4 ,− 2    4 x ……(1) 3 L2: y = –2x + 2 ……(2) By substituting (1) into (2), we have: 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   

11. L1: y =

0 − (−2) 2 =− −3−0 3 2 L1: y – (–2) = − (x – 0) 3 2 y = − x – 2 ……(1) 3 ∵ L2 ⊥ L1 ∴ Slope of L2 × slope of L1 = –1 2 Slope of L2 ×  −= –1   3 3 Slope of L2 = 2 3 L2: y = x ……(2) 2 By substituting (2) into (1), we have:

12. Slope of L1 =

2 3 x=− x–2 3 2 12 x=− 13 By substituting x = −

12 into (2), we have: 13

32

12 Coordinate Treatment of Simple Locus Problems

3  12  −  2  13  18 =− 13

y=

 12 18  ∴ 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 y = –5 ∴ The coordinates of A = (−1, − 5) 0 − (−5) 5 = 2 − (−1) 3 ∴ The equation of L is 5 y – 0 = (x – 2) 3 5 10 y = x− 3 3

(b) Slope of L =

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 3 x=− 2 3 By substituting x = − into (1), we have: 2

15. (a) Slope of L1 =

y=− =

3 +3 2

3 2

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

(b) Slope of L = −

3 2 3 − 2

−1 −0

=−

1 3

∴ The equation of L is 1 y – 1 = − (x – 0) 3 1 y=− 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 (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 ∴ The coordinates of P = (1, − 1) (c) From L3: x – 2y = 0, we have: 1 L3 = − ( −2) Slope of 1 = 2 ∵ L ⊥ L3 ∴ Slope of L × slope of L3 = –1 1 Slope of L × = –1 2 Slope of L = –2 ∴ The equation of L is y – (–1) = –2(x – 1) y = –2x + 1

33

Certificate Mathematics in Action Full Solutions 5A

17. (a) ∵ L1 ⊥ L2 ∴ Slope of L1 × slope of L2 = –1 –2 × slope of L2 = –1 1 Slope of L2 = 2 ∴ The equation of L2 is 1 y – (–4) = (x – 0) 2 1 y= x–4 2 1 Put y = 0 into y = x – 4, we have: 2 1 0= x–4 2 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 1 = 2 1 ∴ The equation of L3 is y = x. 2 (b) From (a), we have: L1: y = –2x + 16 ……(1) 1 and L3: y = x ……(2) 2 By substituting (2) into (1), we have: 1 x = –2x + 16 2 32 x= 5 32 By substituting x = into (2), we have: 5 1  32  y=   2 5  16 = 5  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)2 + (y – 0)2 = 72 ∴ Centre = (0,0) and radius = 7

(b)

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

(c)

(3x – 1)2 + (3y + 2)2 = 54 (3x – 1)2 + [3y – (–2)]2 = 54 2

2

    1   2   3 x −  + 3 y −  −   = 54 3     3     2

2

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

2

2

2

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

3.

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

2

−4  10  Radius =  2  +  2  − 28 =1  (−6) 9  = − ,−  2 2 (b) Centre  9  =  3,−  2  2

2

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

3 13 2

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

34

12 Coordinate Treatment of Simple Locus Problems

The equation of the circle is (x – 3)2 + (y – 2)2 = ( 2 2 ) 2 2 x + y2 – 6x – 4y + 5 = 0

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

2

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

22 2

4.

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

5.

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

6.

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) + [6 − (−2)] 2

2

=4 5 The equation of the circle is (x – 1)2 + [y – (–2)]2 = ( 4 5 ) 2 2 x + 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

8.

9.

= π (3 3 ) 2 = 27π

∵ 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 (a) Radius = (1 − 3) + ( 4 − 2) =2 2

(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

35

Certificate Mathematics in Action Full Solutions 5A

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 2  = (3, − 2)

2

2

−6 4 =   +   − ( −4) radius  2  2 = 17 Distance between the point (8, 5) and the centre = (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 a =0 or a = −3 14. For the circle C: x2 + y2 – 2x + 4y + 8 = 0, 2

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

2

2

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

 (−2) 4  ,−  = (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.  D E Centre =  − ,−   2 2 ∵ The centre lies on the x-axis. E ∴ − =0 2 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.  D E Centre =  − ,−   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.  D E Centre =  − ,−   2 2 ∵ The centre lies on the straight line x + y = 0.

36

12 Coordinate Treatment of Simple Locus Problems

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: –5E + (–5 – E) = –25 10 E= 3 ∴ D = –E 10 =− 3 10 By substituting E = into (2), we have: 3 10 F = –5 – 3 25 =− 3 ∴ The equation of the circle is 10 10 25 x2 + y2 − 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

0−9 −3−0 =3 =

(c) ∵ L ⊥ AP ∴ Slope of L × slope of AP = –1 Slope of L × 3 = –1 1 Slope of L = − 3 The equation of L is

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

−5 − 3 6−2 = –2 −3 − (−5) 1 = Slope of BC = 10 − 6 2

21. (a) Slope of AB =

Slope of AB × slope of BC = –2 ×

1 2

= –1 ∴ 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 ) 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).

(–

4−0 (b) Slope of BP = 0 − (−2) =2 ∵ AQ // BP ∴ Slope of AQ = slope of BP =2 The equation of AQ is y – 0 = 2[x – (–8)]

37

Certificate Mathematics in Action Full Solutions 5A

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

2 − (−3) 8 − ( −2) 1 = 2 ∵ Slope of MN = slope of BC ∴ MN // BC Slope of BC

3.

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

Level 1 (a) (i) Slope of AB

Slope of BC

5−0 3 − ( −2) =1 =

9−5 7−3 =1 =

4.

L : 2x − 3y + 5 = 0 1 3 y = 2x + 5 2 5 y = x+ 3 3 2 Slope of L1 = 3 ∵ L // L1 ∴ Slope of L = slope of L1 2 = 3 The equation of L is 2 y= 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

(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. (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.

 3 + ( −2) 4 + ( −3)  M = ,  2 2   (a) The coordinates of 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

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

Revision Exercise 12 (p. 88)

1.

=

(b) For the straight line L2: x + (3m + 1)y – 4 = 0, 1 slope = − 3m + 1 ∵ L ⊥ L2 ∴ Slope of L × slope of L2 = –1 1   m×−  = −1 3 m + 1  1 m= − 2

38

12 Coordinate Treatment of Simple Locus Problems

6.

7.

8.

∵ L ⊥ L1 ∴ Slope of L × slope of L1 = –1  3 Slope of L ×  −= –1   5 5 Slope of L = 3 The equation of L is 5 y= x+3 3 5x – 3y + 9 = 0 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  2  Slope of L ×  − = –1  (−1)  1 Slope of L = − 2 The equation of L is 1 y – 0 = − [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 5−2 = –1 The required equation is y – 5 = –1(x – 2) x+y–7=0

(b) Slope of PQ =

9.

(a) ∵ A, B and P are on the same straight line. Slope of AP = slope of BP p − 0 0 − ( −4) ∴ = 4 − 1 1 − (−3) p=3

(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 0 − (−4) y − 0 = 1 − ( −3) 0 − 1 y = −1 ∴ The coordinates of Q = (0,−1) 10. 3a = 4b b 3 = a 4 b−0 b 3 =− =− 0−a a 4 The equation of L is 3 y – 3 = − (x – 2) 4 3x + 4y – 18 = 0 Slope of L =

11. Let a be the x-intercept and y-intercept of the straight line where a > 0. a−0 Slope of the straight line = 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 12 − 4 Slope of L × = –1 7−3 Slope of L × 2 = –1 1 Slope of L = − 2  3 + 7 4 + 12  ,  Mid-point of AB =  2   2 = (5, 8) ∵ L bisects AB. ∴ L passes through the mid-point of AB. The equation of L is 1 y – 8 = − (x – 5) 2 x + 2y – 21 = 0 39

Certificate Mathematics in Action Full Solutions 5A

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) ∵ L ⊥ L2 ∴ Slope of L × slope of L2 = –1  2 Slope of L ×  −  = –1  1 1 Slope of L = 2 The equation of L is 1 y – 0 = (x – 2) 2 x – 2y – 2 = 0 14. (a)

2(2 y + 6) + y = 0 5 y + 12 = 0 y=−

By substituting y = −

(b) ∵ L ⊥ OB ∴ Slope of L × slope of OB = –1 1 − × slope of OB = –1 ( −2) 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:

12 into (2), we have: 5

 12  x = 2 −  + 6  5 6 = 5  6 12  ∴ The coordinates of B =  5 , − 5    15. (a) The equation of the circle is x 2 + y 2 = 22 ∴ x2 + y2 − 4 = 0 (b) Radius

= (3 − 0) 2 + (−2 − 1) 2

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

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°)

12 5

x2 + y2 – 2y – 17 = 0

 −2 + 2 3 + (−5)  ,  (c) Centre =  2  2  = (0, –1) 1 Radius = × 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 ∴ x2 + y2 + 4x – 6y + 9 = 0 16. (a) The centre (0, k) is at the same distance from (3, 5) and (5, –1). (0 − 3) 2 + (k − 5) 2 = (0 − 5) 2 + [k − ( −1)] 2 ∴

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

40

12 Coordinate Treatment of Simple Locus Problems

2  = (0 − 3) 2 +  − 5  (b) Radius 3   5 = 10 3 The equation of the circle is

2

2

2  5  ( x − 0) 2 +  y −  =  10  3  3  ∴ 2 4 82 2 x + y − 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 20 D=− 3 20 By substituting D = − into (2), we have: 3 20 − + F = –1 3 17 F= 3 20 17 By substituting D = − and F = into (3), we 3 3 have: 17  20  3 −  + 2 E + = −13 3  3  2 E= 3 ∴ The equation of the circle is 20 2 17 x2 + y2 − x+ y+ = 0. 3 3 3

( ) ()

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

2

5 2 3

19. (a) For the equation C1: x2 + y2 + 4x + 6y – 10 = 0,  4 6 = − ,− centre  2 2  = (−2,−3) (b) ∵ C1 and C2 are concentric. ∴ Centre of C2 = centre of C1 = (–2, –3) 2 2 Radius of C = [5 − (−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 ( x − 0) 2 + (0 − y ) 2 = 10 x2 + y2 = 100 Put x = 6, then y = 8. 0−8 = Slope of L 6 − 0 4 =− 3 ∴

2

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

18. Let the equation of the circle be x2 + y2 + Dx + Ey + F = 0.  D E Centre =  − ,−   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.

41

Certificate Mathematics in Action Full Solutions 5A

∴ A possible equation of L is 4 y=− x+8 3 4x + 3y – 24 = 0 Put x = −8 , then y = 6 . 6−0 Slope of L = 0 − ( −8) 3 = 4 ∴ A possible equation of L is 3 y = x+6 4 3 x − 4 y + 24 = 0 ∴ The possible equation of straight line L is 4 x + 3 y − 24 = 0 or 3 x − 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 6 x-intercept of AB = − = –3 2 (−11) x-intercept of AC = − = 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< m<− or 3 2 Let m = 0. ∴ A possible equation of L is 0 x − y + [4 − 3(0)] = 0 y=4 1 Let m = . 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 1 1 x − y + [4 − 3 ] = 0 3 3 x − 3y + 9 = 0 1 Let m = − . 3 ∴ A possible equation of L is 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 1 Slope of L = − 2 The equation of L is 1 y – 2 = − (x – 6) 2 x + 2y – 10 = 0 ……(1) (c) ∵ 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). 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).

42

12 Coordinate Treatment of Simple Locus Problems

Slope of the required line

4−0 −2−5 4 =− 7 =

The required equation is 4 y – 0 = − (x – 5) 7 4x + 7y – 20 = 0 24. (a) (i) Slope of AB =

5 −1 4 = 1 − (−2) 3

The required equation is 4 y – 1 = [x – (–2)] 3 4x – 3y + 11 = 0 2−5 = (ii) Slope of BC 5 − 1 3 =− 4 The required equation is 3 y – 5 = − (x – 1) 4 3x + 4y – 23 = 0 (iii) Slope of AD = slope of BC 3 =− 4 The required equation is 3 y – 1 = − [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  The coordinates of K  2  3 3 = ,   2 2 25. (a) L: 2x + y – 4 = 0 (−4) x-intercept = − =2 2 (−4) y-intercept = − =4 1 ∴ The coordinates of A and B are (2, 0) and (0, 4) respectively. 4−0 0−2 = –2

(b) Slope of L =

∵ L ⊥ L1 ∴ Slope of L × slope of L1 = –1 –2 × slope of L1 = –1 1 Slope of L1 = 2 The equation of L1 is 1 y – 0 = (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 Slope of L1 ×  −= –1   2 Slope of L1 = 2 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).

43

Certificate Mathematics in Action Full Solutions 5A

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

y = ( 6 − 0) 2 + ( 2 − y ) 2 ∴

= (5 − 3) 2 + (5 − 1) 2 =2 5 27. (a) ∵ 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 k = −3 (c) ∵ AD // BC ∴ Slope of AD = slope of BC = –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    (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 2 = 36 + 4 − 4 y + y 2 4 y = 40 y = 10

∴ The coordinates of A = (0,10) 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 3 y – 2 = (x – 6) 4 3x – 4y – 10 = 0 ……(1) (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, (−10) 10 = x-intercept = − 3 3 10 ∴ The coordinates of B = ( , 0). 3 Area of quadrilateral OAPB = 2 × area of OAB 1 = 2 × × OB × OA 2  1  10   = 2 ×  ×  − 0  × (10 − 0) 2 3     =

100 3

29. (a) By substituting (4, 3) into x2 + y2 + 2x + ky – 15 = 0, we have: 42 + 32 + 2(4) + k(3) – 15 = 0 3k = –18 k = −6

44

12 Coordinate Treatment of Simple Locus Problems

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

2

2 −6 Radius =  2  +  2  − (−15)     =5 The distance between B and the centre of circle C = [−1 − (−6)]2 + (3 − 3) 2 =5 = radius of circle C ∴ The point B lies on the circle.  4 + ( −6) 3 + 3  ,  (c) Mid-point of A and B =  2 2   = (–1, 3) = centre of circle C Also, A and B lie on the circle. ∴ AB is a diameter of the circle C.  4 ( −6)  30. (a) Centre =  − 2 ,− 2  = (−2, 3) 2

2

4 −6 Radius =  2  +  2  − ( −3)     =4 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. (b) Let P be the centre of the circle C. 4−3 = Slope of PM 1 − (−2) 1 = 3 ∵ PM ⊥ AB (line joining centre to mid-pt. of chord ⊥ chord) ∴ Slope of PM × slope of AB = –1 1 × slope of AB = –1 3 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) 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 (d) Radius =  2  +  2  − 0     =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, (−24) x-intercept = − 3 =8 (−24) y-intercept = − 4 =6 ∴ The coordinates of A and B are (8, 0) and (0, 6) respectively. (b) Let the equation of the circle be:

45

Certificate Mathematics in Action Full Solutions 5A

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. 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)

(c) tan ∠OBA

(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 (a – 4)(a + 4) = 0 a=4

=

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.

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) (d)

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

=

 AB     2 

AP 6   2

5 3 = 5 ∴ θ = 36.87° ∠APB = 2θ = 2 × 36.87° = 73.74° (a) The equation of the circle is (x – 0)2 + (y – 28)2 = 202 x2 + y2 – 56y + 384 = 0 ……(1)

Area of sector PAKB = π × 52 ×

73.74° 360°

= 16.09

46

12 Coordinate Treatment of Simple Locus Problems

1 × AP × BP × sin 73.74° 2 1 = × 5 × 5 × sin 73.74° 2 = 12.00 ∴ Area of segment AKB = area of sector PAKB – area of PAB = 16.09 – 12.00 = 4.1 (cor. to 1 d.p.)

∴ The coordinates of C are (6, 5). 1 ∴ Area of ABC = × AB × AC 2 1 = × (5 – 2) × (6 – 0) 2 =9

Area of PAB =

3.

Answer: C Let L be the line joining (–3, 2) and (4, –1). −1 − 2 = Slope of L 4 − (−3) 3 =− 7 7 The required line should have slope . 3

4.

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).

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

2

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

2

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

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

(b) Distance between the two centres = ( 2 − 6) 2 + [1 − (−2)]2 =5 =2+3 = radius of C1 + radius of C2 ∴ The two circles touch each other.

∴ AC

5.

= (7 − 2) 2 + (1 − 4) 2 = 34

Answer: D Slope of L1 = − 1 2

Multiple Choice Questions (p. 94)

Slope of L2 =

1.

Slope of L3 = −

2.

Answer: D ∵ The two lines are parallel. ∴ They have the same slope. 2 1 − =− (−k ) 3 k = –6 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 − 2y + 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 x=6

4 =2 ( −2)

4 = –2 ≠ 2 2 ∴ L1 is not parallel to L3. ∴ B is false. 1 = 1 ≠ –1 2 ∴ L1 is not perpendicular to L2. ∴ A is false. Slope of L1 × slope of L2 = 2 ×

Slope of L1 × slope of L3 = 2 × –2 = –4 ≠ –1 ∴ L1 is not perpendicular to L3. ∴ C is false. ( −5) 5 = x-intercept of L1 = − 4 4 ( −5) 5 = x-intercept of L3 = − 4 4 x-intercept of L1 = x-intercept of L3 5 ∴ L1 and L3 intersect at the point ( , 0) which is on the 4 47

Certificate Mathematics in Action Full Solutions 5A

x-axis. 6.

Answer: D c <0 a c y-intercept of L = − < 0 b a Slope of L = − < 0 b x-intercept of L = −

7.

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

8.

Answer: A

∵ Diagonals bisect each other. ∴ The mid-point P of A and C lies on BD.  −4 + 6 10 + (−4)  ,  = (1, 3) P = 2  2  3− 2 1 = Slope of BD = slope of BP = 1 − ( −8) 9 Only equation of option A has the same slope. 9.

Answer: C From the figure, x-intercept of L < 0. c ∴ − <0 1 c>0 y-intercept of L > 0 c ∴ − >0 b b<0

10. Answer: A

 k (−8)   k  Centre of C =  − 2 ,− 2  =  − 2 ,4      Since L divides C into two equal parts, L passes through the centre of C.  k  By substituting  − 2 , 4  into the equation of L, we have:    k 2  −  +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 For A, radius =   +  2  2  ∴ A represents an imaginary circle. 2

1 2  − 2    2   1 =  −  +  2   2  2 For D, radius 3 = >0 4 ∴ 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  (−6) 10  ,−  Centre of C =  − 2 2  = (3, –5) ∴ I is false. 2

2

Radius =  − 6  +  10  − 9  2   2 =5 ∴ II is true. The distance between (–1, –1) and the centre

48

12 Coordinate Treatment of Simple Locus Problems

= (−1 − 3 ) 2 + [ −1 − ( −5)]2 =4 2 >5 = radius of circle C ∴ (–1, –1) lies outside the circle. ∴ III is true. 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.  (−6) 4  ,−  Centre of C =  − 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.

49