146
Chapter 10 Straight Lines and Rectilinear Figures
(d) Let (x, y) be the coordinates of the point.
CHAPTER 10 Exercise 10A (p.233)
x=
1. (a) d = ( 4 − 2)2 + (7 − 1)2 = 40 = 2 10
0 + 12 ⋅ 3 1+
1 2
= 1, y =
3 + 12 ⋅ 0 1+
1 2
=2
∴ The point is (1, 2) .
(b) d = [ −5 − ( −2)] + [3 − ( −1)] = 5 2
2
(c) d = (2 p − p)2 + ( − q − q )2 =
p2 + 4q 2
(d) d = ( − a − a)2 + [ − a − ( −2 a)]2 = 5a 2.
( p + 5)2 + (8 − 3)2 = 13
−1 + 7 4 + 0 , ) = (3, 2) 2 2 3+3 2+6 N =( , ) = (3, 4) 2 2
6. M = (
7. Let (x, y) be the coordinates of the point.
2
When r = 2 ,
( p + 5) + 25 = 169 ( p + 5) = 144 p + 5 = ±12 2
−4 + 2(5) =2 1+ 2 −1 + 2(11) y= =7 1+ 2 1 When r = , 2 x=
p = 7 or −17 3. Let (2a, a) be the coordinates of P. (2 a − 3)2 + ( a − 1)2 = (2 a − 5)2 + ( a − 3)2 5a 2 − 14 a + 10 = 5a 2 − 26 a + 34 12 a = 24 a=2 ∴ The point P is ( 4, 2) .
x= y=
−4 + 12 (5) 1+
1 2 −1 + 12 (11) 1 + 12
= −1 =3
∴ The points are (2, 7) and (−1, 3).
4. Let (0, y) be the coordinates of C. AB = BC [ −3 − ( −2)]2 + (2 − 4)2 = [0 − ( −2)]2 + ( y − 4)2 12 + 2 2 = 2 2 + ( y − 4)2 ( y − 4)2 = 1 y = 4 ±1 = 3 or 5 ∴ C is (0, 3) or (0, 5).
5. (a) Let (x, y) be the coordinates of the point. −1 + 4 ⋅ 4 15 −3 + 4 ⋅ 2 5 = = 3, y = = =1 1+ 4 5 1+ 4 5 ∴ The point is (3, 1) . x=
8. Let AP : PB = r : 1 and (0, y) be the coordinates of P. −3 + 2 r =0 1+ r 3 r= 2 3 ∴ AP : PB = : 1 = 3 : 2 2 y=
2 + 3(3) 11 = 3+2 5
∴ The point P is (0,
(b) Let (x, y) be the coordinates of the point. x=
−1 − 14 ⋅ 2 1−
1 4
= −2 , y =
1 − 14 ⋅ 1 1−
1 4
=1
∴ The point is ( −2, 1) . (c) Let (x, y) be the coordinates of the point. 3− 2⋅5 1− 2⋅3 = 7, y = =5 1− 2 1− 2 ∴ The point is (7, 5) . x=
9. Let AP : PB = r : 1 −1 + 3r =2 r +1 r=3 ∴ AP : PB = 3 : 1 2 + 6(3) 3 +1 =5
k=
11 ). 5
Chapter 10 Straight Lines and Rectilinear Figures
10. Let (x, y) be the coordinates of the other end of this diameter. 3+ x =4 2 x=5 −4 + y =5 2 y = 14 ∴ The other end is (5, 14) . m +1+1 2 −1 , ) 2 2 m+2 1 =( , ) 2 2
11. (a) Mid-point of AC = (
−2 + 3 −2 + n + 3 , ) 2 2 1 n +1 =( , ) 2 2
Mid-pint of BD = (
m+2 1 (b) = 2 2 m = −1
0 −1 1 4 14. Area = 2 3 0
2 0 0 square units 3 2
1 (12 + 6 + 2) square units 2 = 10 square units
=
15. (2t + 2)2 + t 2 = t 2 + (3t − 3)2 (2t + 2)2 − (3t − 3)2 = 0 (2t + 2 + 3t − 3)(2t + 2 − 3t + 3) = 0 (5t − 1)( −t + 5) = 0 5t − 1 = 0 or −t + 5 = 0 1 t = or t = 5 5 16. (a) AB = (5 − 1)2 + (9 − 2)2 = 65 AC = (5 − 4)2 + (9 − 1)2 = 65 ∴ AB = AC (b) Q BP = CP
n +1 1 = 2 2 n=0
∴
(1 − h)2 + (2 − k )2 = ( 4 − h)2 + (1 − k )2
1 − 2h + h 2 + 4 − 4k + k 2 = 16 − 8h + h 2 + 1 − 2 k + k 2 6h − 2 k − 12 = 0 3h − k − 6 = 0
3( 4) + 2 a 12. =2 3+2 a = −1 2( 4) + 3( −1) =b 3+2 b =1
5 5 1 −6 7 13. (a) Area = 2 −7−2 5 5
147
17. (a) BC = (5 + 1)2 + ( −4 + 1)2 = 6 2 + 32 = 3 5 (b) Area of ∆ABC
square units
1 = (35 + 12 − 35 + 30 + 49 + 10) square 2 units = 50.5 square units 3 0 1 4 3 (b) Area = square units 2 −2 −1 3 0 1 = (9 − 4 + 6 + 3) square units 2 = 7 square units
3 2 1 −1 −1 square units = 2 5−4 3 2 1 ( −3 + 4 + 10 + 2 + 5 + 12) square units 2 = 15 square units
=
(c) Length =
2 ⋅ 15 =2 5 3 5
2 −1 1 a +1 a − 3 18. Area = square units a 2 a+2 2 −1 1 = a − square units 2
148
Chapter 10 Straight Lines and Rectilinear Figures
Exercise 10B (p.240)
1 , area = 0 2 ∴ They are collinear. When a =
1. Let m be the slope and α be the inclination. Then 6−0 3 = 4−0 2 α = 56.3° (corr. to 1 d.p.) 3 ∴ Slope = 2
(a) m = tan α = 19. Let (a, 0) be the coordinates of C. a 0 1 5 −1 ∴ Area of ∆ABC = 2 −2 2 a 0
∴
square units
Inclination = 56.3° 1 − ( −4) =1 6 −1 α = 45° (corr. to 1 d.p.) ∴ Slope = 1 Inclination = 45°
(b) m = tan α =
8 − 3a = 20 8 − 3a = 20
8 − 3a = −20 28 a = −4 or a= 3 28 ∴ C is ( −4, 0) or ( , 0) . 3 or
2r 2 , 2) , Q(2, ) 1+ r 1+ r (b) Area of ∆OPQ
20. (a) P(
2 1+2 r 2r 1 1+ r 2 = 2 0 0 2 1+2 r
∴ The y-intercept is −3 .
square units
=
1 4r square units 4− 2 (1 + r )2
=
1 4(r 2 + r + 1) square units 2 (1 + r )2
As r > 0 ,
2. Let C(0, b) and D(a, 0) be the intersections of AB with the coordinate axes. Then Slope of AC = slope of AB −4 − b −4 − 3 = −1 − 0 −1 − 6 4+b =1 b = −3
r2 + r + 1 > 0. (1 + r )2
Slope of AD = slope of AB −4 − 0 −4 − 3 = −1 − a −1 − 6 4 = 1+ a a=3
∴ The x-intercept is 3 .
2 −1 1 = 1 − ( −3) 4 4−2 1 Slope of BC = = 9 −1 4 ∴ AB // BC They are parallel and have a common point B. Therefore A, B and C are collinear.
3. Slope of AB =
∴ Area of ∆OPQ =
2(r 2 + r + 1) square units (1 + r )2
r2 + r + 1 (1 + r )2 ∴ y(1 + 2 r + r 2 ) = r 2 + r + 1 (1 − y)r 2 + (1 − 2 y)r + (1 − y) = 0
(c) Let y =
As r is real, ∴ (1 − 2 y)2 − 4(1 − y)2 ≥ 0 3 y≥ 4 3 3 ∴ Area of ∆OPQ = 2 y ≥ 2( ) = 4 2 ∴ Minimum area of ∆OPQ 3 = square units 2
4.
2 − ( −4) −7 − 2 = −7 − 1 x − ( −7) 6 −9 = −8 x + 7 x=5
5. Let the acute angle be α. 6 −1 5 = 1 + (6)(1) 7 α = 35.5° (corr. to 1 d.p.)
(a) tan α =
Chapter 10 Straight Lines and Rectilinear Figures
0.6 − 0.25 7 = 1 + (0.6)(0.25) 23 α = 16.9° (corr. to 1 d.p.)
(b) tan α =
4 − ( − 23 )
14 (c) tan α = = 5 1 + 4( − 23 ) α = 70.3° (corr. to 1 d.p.) −1 − ( − 43 )
1 (d) tan α = = 4 7 1 + ( −1)( − 3 ) (corr. to 1 d.p.) α = 8.1° (e) m1 = m2 = − ∴
α = 0°
1 2
6. Slope of AB =
5 −1 1 = 6+2 2
Slope of CD =
0+2 1 = 4−0 2
∴
AB ⊥ AC
(1 + 5)2 + ( 4 − 6)2 square units 1 ( 10 )(2 10 ) square units 2 = 10 square units
=
10. Let A = (−2, −1), B = (0, 2), C = (3, 0), D = (1, −3)
5 +1 3 =− −2 − 2 2
1+1 2 = 5−2 3 ∴ (Slope of QR) ⋅ (Slope of PQ) = −1 PQ ⊥ QR y
8.
7−4 =3 2 −1 7−6 1 Slope of BC = = 2+5 7 4−6 1 Slope of CA = =− 1+ 5 3 ∴ (Slope of AB)(Slope of CA) = −1
9. Slope of AB =
1 ( AB)( AC ) 2 1 = (7 − 4)2 + (2 − 1)2 2
Slope of QR =
∴
A = 180° − B − C = 180° − 53.1° − 45° = 81.9° (corr. to 1 d.p.)
Area =
AB // CD
7. Slope of PQ =
−1 − 0 =1 1 C = 45°
tan C =
∴
1 (f) m1m2 = ( −2)( ) = −1 2 ∴ α = 90°
A
AB = BC = CD = DA = 2 2 + 32 = 13 3 Slope of AB = Slope of CD = 2 2 Slope of BC = Slope of DA = − 3 (Slope of AB)(Slope of BC) = −1 (Slope of CD)(Slope of DA) = −1 ∴ A, B, C, D form a square area Area = ( 13 )2 square units = 13 square units
B −4
−1 0
149
C 3
x
4−0 4 = −1 + 4 3 Slope of BC = 0 4−0 Slope of AC = = −1 −1 − 3 Slope of AB =
4 −0 4 tan B = 3 = 1 3
B = 53.1° (corr. to 1 d.p.)
11. Let (a, 0) be the coordinates of C. −3 + 2 1 Slope of AB = = −1 − 2 3 −3 − 0 3 Slope of BC = = −1 − a 1 + a Q AB ⊥ BC ∴ (Slope of AB)(Slope of BC) = −1 1 3 ( )( ) = −1 3 1+ a 1 = −1 − a a = −2 ∴ The point C is ( −2, 0) .
150
Chapter 10 Straight Lines and Rectilinear Figures
(b) BD : DA = OB : OA = 2 : 4 = 1 : 2 Let (x, y) be the coordinates of D.
0−2 k−2 = h −1 0 −1 (h − 1)( k − 2) = 2 2 h −1 = k−2 k h= k−2
12. Slope =
13. Let A be (1, 0) , B be (t 2 , 2t ) , C be ( 2t − 0 2t = Slope of AB = 2 t − 1 t2 − 1
Slope of CA =
∴
∴ The coordinates of D is (2, 1 2 , − ). 2 t t
(c) (i) Slope of OD = Slope of OB =
2 t
2t = 1 − 12 t 2 − 1 t
∴ AB // CA They are parallel and have a common point A. Therefore A, B, C are collinear. 14. (a) BC = 2 a ∴ The coordinates of C = (2 a cos θ + a, 2 a sin θ) DA = BC = 2 a ∴ The coordinates of D = (2 a cos θ, 2 a sin θ)
(b) Coordinates of F = (2 a, 0) Coordinates of E = ( − a, 0) 2 a sin θ Slope of DF = 2 a cos θ − 2 a sin θ = cos θ − 1 2 a sin θ Slope of CE = 2 a cos θ + 2 a sin θ = cos θ + 1 (Slope of DF)(Slope of CE) sin θ sin θ =( )( ) cos θ − 1 cos θ + 1 sin 2 θ = cos 2 θ − 1 sin 2 θ = − sin 2 θ = −1 ∴ DF ⊥ CE 15. (a) OA = 4 OB = 12 + ( 3 )2 = 2
1(2) + 4(1) 6 = =2 1+ 2 3 3 (2) + 0(1) 2 3 y= = 1+ 2 3 x=
(ii)
2 3 3
−0
2−0
=
2 3 ). 3
3 3
3 = 3 1
3 ) = 30° 3 −1 ∠BOA = tan ( 3 ) = 60° ∴ ∠BOD = 60° − 30° = 30° = ∠DOA ∴ OD is the angle bisector of ∠AOB . ∠DOA = tan −1 (
Exercise 10C (p.246) 1. (a) y − 2 = 3( x − 1) y − 2 = 3x − 3 3x − y − 1 = 0 1 (b) y − 3 = − ( x − 2) 2 2y − 6 = −x + 2 x + 2y − 8 = 0 (c) y − 0 = −4( x + 1) y = −4 x − 4 4x + y + 4 = 0 (d) y − 1 = 0( x − 2) y −1 = 0 4 −1 ( x − 3) 2−3 y − 1 = −3 x + 9 3 x + y − 10 = 0
2. (a) y − 1 =
−2 − 0 ( x − 5) −1 − 5 3y = x − 5 x − 3y − 5 = 0
(b) y − 0 =
−4 − 3 −7 = and the 1−1 0 x-coordinates of the points are 1.
(c) Q The slope = ∴
x −1 = 0
151
Chapter 10 Straight Lines and Rectilinear Figures
y-intercept is −2 .
4a − 2a ( x − a) 3a − a y − 2a = x − a x−y+a=0
(d) y − 2 a =
There is no x-intercept. y
−2
3. (a) x + 3 y − 9 = 0 3y = − x + 9 1 y = − x+3 3 1 ∴ The slope is − . 3
x
−1 O −1 −2
1
2
y+2=0
(d) 2 x + 5 = 0 5 2 ∴ The slope is undefined. There is no y-intercept. x=−
y-intercept is 3 . Substitute y = 0 into the equation. ∴
x−9=0 x=9 ∴ x-intercept is 9 .
5 x-intercept is − . 2 y
y 2x + 5 = 0
3
5 2
1
−2
−1 O −1
−
x + 3y − 9 = 0
2
−4
−3
1
x
O
1
2
3
4
5
6
7
8
9
x + 3 y − 11 = 0 4. (a) 2 x − y −1 = 0
(b) 2 x − 5 y + 10 = 0 5 y = 2 x + 10 2 y= x+2 5 2 ∴ The slope is . 5
7 y − 21 = 0 y=3 x=2
The intersection is (2, 3) .
y-intercept is 2 .
4 x − 5 y − 12 = 0 (b) x − 2 y − 6 = 0
Substitute y = 0 into the equation. ∴
2 x + 10 = 0 x = −5 ∴ x-intercept is −5 .
3 y + 12 = 0 y = −4 x = −2
The intersection is ( −2, − 4) .
y
2x − 5y + 10 = 0
2
5. Perpendicular bisector of BC: 1
−5
−4
−3
−2
(c) y + 2 = 0 y = −2 ∴ The slope is 0 .
−1 O
x
[
y− x−
1+ ( −3) 2 ][ 1 − ( −3) ] = 7 + ( −1) 7 − ( −1) 2
−1
y +1 = −2 x−3 y + 1 = −2 x + 6 2x + y − 5 = 0
x
152
Chapter 10 Straight Lines and Rectilinear Figures
Perpendicular bisector of AC: y − 62+1 6 − 1 ( )( ) = −1 x − 7 +2 2 2 − 7 7 9 y− = x− 2 2 x − y −1 = 0
2 x + y − 5 = 0 x − y − 1 = 0 3x − 6 = 0 x=2 y =1 ∴ The circumcentre is (2, 1) . 6. (a) x − 3 y + 4 = 0 3y = x + 4 1 4 y= x+ 3 3 1 ∴ Slope is . 3 2x − y − 5 = 0 y = 2x − 5 ∴ Slope is 2. Let α be the acute angle. tan α =
2 − 13 1 + 2( 13 )
=1
α = 45° (b) 2 x + 5 y + 4 = 0 5 y = −2 x − 4 2 4 y=− x− 5 5 2 ∴ Slope is − . 5 x − 3y = 0 3y = x 1 y= x 3 1 ∴ Slope is . 3 Let α be the acute angle. tan α =
− 25 − 13
=
11 13
1 + (− α = 40.2° (corr. to 3 sig. fig.) 2 )( 1 ) 5 3
7. Let A and B be (a, 0) and (0, b) respectively. If OA = OB , then a = b or a = − b . i.e. m = 1
or
m = −1
y −1 = x − 4
or
y − 1 = −( x − 4 )
x − y − 3 = 0 or
x+y−5=0
8. (a) 3 x − 4 y − 1 = 0 4 y = 3x − 1 3 1 y= x− 4 4 3 Slope is . 4 ∴ The required equation: 3 y − 2 = ( x − 5) 4 3 x − 15 = 4 y − 8 3x − 4 y − 7 = 0 (b) Slope of required equation =
−1 3 4
=−
4 3
∴ The required equation: 4 ( x − 5) 3 −3 y + 6 = 4 x − 20 4 x + 3 y − 26 = 0 y−2= −
9. 4 x + 6 y − 3 = 0 6 y = −4 x + 3 2 1 y=− x+ 3 2 2 Slope is − . 3 3 x + ky − 7 = 0 ky = −3 x + 7 3 7 y=− x+ k k 3 Slope is − . k Q Two lines are perpendicular to each other. ∴
2 3 ( − )( − ) = −1 3 k 2 = −1 k k = −2
5x − y − 7 = 0 10. 18 x − 7 y + 2 = 0 17 x − 51 = 0 x=3 Substitute x = 3 into L1 , 5(3) − y − 7 = 0 y=8 The intersection is (3, 8).
Chapter 10 Straight Lines and Rectilinear Figures
By using the two-point form, the equation of the line is 8−2 y−2= ( x − 1) 3 −1 2y − 4 = 6x − 6 6x − 2y − 2 = 0 3x − y − 1 = 0 11. Slope of CD =
153
Substitute y = −2 into AD, 2x − 2 − 6 = 0 x=4 ∴ The coordinates of D are ( 4, − 2) . 13. x-intercept of 2 x − y + 4 = 0 is −2. x-intercept of 4 x + y − 4 = 0 is 1. 2 x − y + 4 = 0 4 x + y − 4 = 0
2 5 −1
5 Slope of AD = 2 = − 2 5
By using point-slope form, the equation of AD is 5 y − 5 = − ( x − 2) 2 −2 y + 10 = 5 x − 10 5 x + 2 y − 20 = 0 2 5 By using point-slope form, the equation of AB is 2 y − 5 = ( x − 2) 5 2 x − 4 = 5 y − 25 2 x − 5 y + 21 = 0
Slope of AB = slope of CD =
6x = 0 x=0 y=4
The intersection is (0, 4). 0 1 −2 Area = 2 1 0
4 0 square units 0 4
1 ( 4 + 8) square units 2 = 6 square units =
14. Let ( s, 1 − s) be the coordinates of the required point. ( s − 2)2 + (1 − s − 2)2 = ( s − 3)2 + (1 − s + 1)2
12. (a) A is the intersection of AB and CA. 3 x − y − 4 = 0 3 x + 4 y − 14 = 0 5 y − 10 = 0 y=2 Substitute y = 2 into AB, 3x − 2 − 4 = 0 x=2 ∴ The coordinates of A are (2, 2) . (b) Slope of BC = Slope of AD =
1 2 −1 1 2
s 2 − 4s + 4 + s 2 + 2 s + 1 = s 2 − 6s + 9 + s 2 − 4s + 4 8s = 8 s =1 ∴ The required point is (1, 0) .
15. Slope of 4 x − y − 5 = 0 is 4. Slope of ax − 15 y + 7 = 0 is tan 45° =
= −2
By using point-slope form, the equation of AD is y − 2 = − 2( x − 2 ) y − 2 = −2 x + 4 2x + y − 6 = 0 D is the intersection of AD and BC. 2 x + y − 6 = 0 x − 2 y − 8 = 0 5 y + 10 = 0 y = −2
1=
a 15
a . 15
−4
a) 1 + ( 4)( 15
a − 60 15 + 4 a
a − 60 a − 60 = 1 or = −1 15 + 4 a 15 + 4 a or 5a = 45 3a = −75 a = −25 or a = 9 16. (a) By using point-slope form, the equation of L is y−4 =m x−2 y − 4 = m( x − 2 )
154
Chapter 10 Straight Lines and Rectilinear Figures
4 +2 m y-intercept of L = −2 m + 4
(b) x-intercept of L = −
Area =
1 −4 ( + 2)( −2 m + 4) square units 2 m
1 −4 ( + 2)( −2 m + 4) 2 m 16 4 = 16 − − 4m m
1 3 The coordinates of M are ( , ) . 2 2 Slope of AB = −1 −1 ∴ Slope of the required line = =1 −1 ∴ The equation of the line is
y − 23
2=
16 − =4
16 − 4m m
or 16 − = −4
=1 x − 12 x − y +1 = 0 16 − 4m m
16 m − 16 − 4 m 2 or 16 m − 16 − 4 m 2 = 4m = −4 m
Exercise 10D (p.258) y
1. (a)
4 m 2 − 12 m + 16 or 4 m 2 − 20 m + 16 = 0 =0 ( m − 1)( m − 4) = 0 no solution ∴ m = 1 or 4
or m = 1 or m = 4 L1
2
135°
x
O
17. Slope of L = 2 In the figure, the equation of L 1 is
Let m be the slope of L1 or L 2 . Then
1 m−2 tan θ = = 3 1 + 2m m−2 1 m−2 1 =− = or 1 + 2m 3 1 + 2m 3 m =1 m = 7 or By using point-slope form,
x cos 135° + y sin 135° − 2 = 0 x y − + −2=0 2 2 x−y+2 2 =0 y
(b)
L1 : 7 x − y + 9 = 0 , L 2 : x − y + 3 = 0 18. 3 x 2 + y 2 = 6 ......................(1) x + y = 2 ...........................(2) From (2), y = 2 − x ...........(3) Substitute (3) into (1), 3 x 2 + (2 − x ) 2 = 6 2x2 − 2x − 1 = 0 Let A, B be ( x 1, y 1 ) and ( x 2 , y 2 ) respectively.
∴
Q ∴ ∴
2 =1 2 1 = 2 =2 =2 y + y2 + 1 =2 2 1 3 =2− = 2 2
x1 + x2 = x1 + x2 2 x1 + y1 x2 + y2 x1 + x2 2 y1 + y2 2
4
L2
30°
O
x
In the figure, the equation of L 2 is
x cos 30° + y sin 30° − 4 = 0 1 3 x+ y−4=0 2 2 3x + y − 8 = 0
Chapter 10 Straight Lines and Rectilinear Figures
2. (a) The normal form is 5 x + 12 y
d1 =
=0
52 + 12 2 5 x + 12 y =0 13 5 12 x+ y=0 13 13
d2 =
7. r =
p=3
(d) d =
2 − ( −3) 2
2
=
5 = 5 5
4 + 21 18 + 24 2
2
=
25 5 = 30 6
3( −1) − 4(5) − 12 32 + 4 2
=7
= 5 12 + 2 2 k −3 = 5
k − 3 = 5 or k − 3 = −5 k = 8 or k = −2 9. Let P(x, y) be a point on the perpendicular bisector, then the distance from P to the two lines are equal.
p = 10
(a)
3(1) + 4( 4) − 9 32 + 4 2
(b)
12 2 + 52 5 − ( −2) − 3 2
2( −4) + 3 0 +2
2
=4
=
2x − y − 5 x + 2y − 2 = 5 5 2 x − y − 5 = ±( x + 2 y − 2) 2 x − y − 5 = x + 2 y − 2 or 2 x − y − 5 = −( x + 2 y − 2 )
=2 2
1 +1 2
x+ y−3 x − y +1 = 2 2 x + y − 3 = ±( x − y + 1) x + y − 3 = − ( x − y + 1) or x + y − 3 = x − y + 1 or x −1 = 0 y−2=0
=2
12( −2) − 5(7) + 7
2
3 5
1 − 2( 2 ) + k
3 1 , sin θ = 2 2 θ = 120°
cos θ = 1 , sin θ = 0 ∴ θ = 0°
(c) d =
=
2
8. By considering the distance from (1, 2) to the line,
(b) The normal form is x − 10 =0 1 x − 10 = 0
(b) d =
2
d=
cos θ = −
4. (a) d =
3 5
2 +1 (b) First rewrite equations as 18 x − 24 y − 21 = 0 , 18 x − 24 y + 4 = 0 .
x 3 − + y−3= 0 2 2
∴
=
=0
− 12 + ( 3 )2
∴
2
4(0) + 3(1) − 6
6. (a) d =
3. (a) The normal form is
∴
3 +4 2
4 +3 ∴ The point (0, 1) is equidistant from the lines.
(b) The normal form is 2x − y − 5 =0 2 2 + 12 2 1 x− y− 5 =0 5 5
x − 3y + 6
3(0) − 4(1) + 1
155
5 2
5. Let d 1 be the distance from (0, 1) to the line 3 x − 4 y + 1 = 0 , d 2 be the distance from (0, 1) to the line 4 x + 3 y − 6 = 0 .
x − 3 y − 3 = 0 or 3 x + y − 7 = 0 (c)
3x + 2 y + 7 2 x − 3y − 1 = 13 13 3 x + 2 y + 7 = ±(2 x − 3 y − 1) or 3 x + 2 y + 7 3x + 2 y + 7 = −(2 x − 3 y − 1) = 2 x − 3y − 1 x + 5 y + 8 = 0 or 5 x − y + 6 = 0
156
Chapter 10 Straight Lines and Rectilinear Figures
(d)
7x − y + 4 2x + 2y − 5 = 5 2 2 2 14 x − 2 y + 8 = ±(10 x + 10 y − 25)
14 x − 2 y + 8 = 10 x + 10 y − 25 or 14 x − 2 y + 8 = −(10 x + 10 y − 25) 4 x − 12 y + 33 = 0 or 24 x + 8 y − 17 = 0
10. (a)
k−6
=4
5 + 12 k − 6 = 52 2
2
k − 6 = 52 or k = 58 or
k − 6 = −52 k = −46
(b) Let the equation of L be 5 x + 12 y + c = 0 . c−6 =5 13 c − 6 = 65
c − 6 = 65 or c − 6 = −65 c = −59 c = 71 or ∴ The equation of L is 5 x + 12 y + 71 = 0 or 5 x + 12 y − 59 = 0 .
2 x − 3y + 1 = 0 13. x + 2 y − 3 = 0
7y − 7 = 0 y =1 x =1 Intersection of the given lines is (1, 1). Slope = 3, y − 1 = 3 ( x − 1) 3x − y − 2 = 0
Passing through (2, −1), −1 − 1 y −1 = ( )( x − 1) 2 −1 y − 1 = −2 x + 2 2x + y − 3 = 0 The slopes of these lines are 3 and −2.
3 − ( −2) =1 1 + 3( −2) θ = 45°
tan θ =
4 and θ is acute. 3 4 3 sin θ = , cos θ = 5 5
14. tan θ = ∴
y θ
11. (3 + k ) x + (2 k − 1) y = 5k + 1 (3 x − y − 1) + k ( x + 2 y − 5) = 0
3 x − y − 1 = 0 x + 2 y − 5 = 0 7x − 7 = 0 x =1 y=2
θ
θ
O
12. Let (x, y) be the coordinates of the point. 3x + y − 2 = 10 10 3 x + y − 2 = ±10
3 x + y − 2 = 10
or
3 x + y − 2 = −10
3 x + y − 12 = 0
or
3x + y + 8 = 0
3 x + y − 12 = 0 or 3 x + y + 8 = 0 2 x − y − 3 = 0 2 x − y − 3 = 0 or 5 x + 5 = 0 5 x − 15 = 0 or x=3 x = −1 y=3 or y = −5 ∴ The point is (3, 3) or (−1, −5).
1
A C
∴ The point P is (1, 2) .
B
2 π −θ 2
x
The equation of AB is π π x cos θ( − θ) + y sin( − θ) − 2 = 0 2 2 x sin θ + y cos θ − 2 = 0 4 3 x+ y−2=0 5 5 4 x + 3 y − 10 = 0 The equation of BC is x cos( π − θ) + y sin( π − θ) − 1 = 0 − x cos θ + y sin θ − 1 = 0 3 4 − x + y −1 = 0 5 5 ∴ 3x − 4 y + 5 = 0
4 x + 3 y − 10 = 0 .................(1) 3 x − 4 y + 5 = 0 ..................(2) 25 y − 50 = 0 y=2 Substitute y = 2 into (1), x = 1 . ∴ The coordinates of B are (1, 2) .
Chapter 10 Straight Lines and Rectilinear Figures
2x − y − 5 = 0 15. x − 3 y − 10 = 0 5 y + 15 = 0 y = −3 x =1
3 , the required equation is 2 6 x + 8y − 3 = 0 .
When c =
3 When c = − , the required equation is 2 6 x + 8y + 3 = 0 .
Intersection of L 1 and L 2 is (1, −3). (a) Slope = 3, y + 3 = 3 (x − 1) y + 3 = 3x − 3 3x − y − 6 = 0
17. (a) An equation of L is y − 4 = k ( x − 1) y = k ( x − 1) + 4 , where k is real. (b) (i) Substitute (3, 7) into L,
(b) Slope of L 1 = 2 Slope of required equation = −
1 2
7 = k (3 − 1) + 4 3 k= 2 ∴ The equation is
1 y + 3 = − ( x − 1) 2 x − 1 = −2 y − 6 x + 2y + 5 = 0
3 ( x − 1) + 4 2 2 y = 3x − 3 + 8 y=
−3 + 3 (c) y + 3 = ( x − 1) −1 − 1 y+3= 0
3x − 2 y + 5 = 0
x−2=0 (d) 2 x+y−2=0 x = 2 , y = −2 −2 + 3 y+3= ( x − 1) 2 −1 x−y−4=0
(ii) The slope of line 3 x − 5 y + 8 = 0 is 3 k ( ) = −1 5 5 k=− 3 The equation is
∴
16. (a) An equation of L is 3 x + 4 y − c = 0 , where c is real. (b) (i) Substitute (−2, 1) into L, 3( −2) + 4(1) − c = 0 c = −2 ∴ The required equation is 3x + 4 y + 2 = 0 .
k−2 π = 4 1 + k ( 12 ) 1 k 1 + = ±( k − ) 2 2 1 k = 3 or − 3 When k = 3 , the equation is 1
3
(iii) The distance of L from origin is 3 = 10 32 + 4 2 3 c=± 2
1 . 2
tan
4
∴ The sum of intercepts is c + c = 21 3 4 4 c=9 The required equation is 3 x + 4 y − 9 = 0 .
c
5 ( x − 1) + 4 3 3 y = − 5 x + 5 + 12 5 x + 3 y − 17 = 0 y=−
(iii) Slope of x − 2 y + 1 = 0 is
(ii) Express L in intercept-form: 3 x + 4 y = c x y + c =1 c
∴
157
y = 3( x − 1) + 4 . ∴
3x − y + 1 = 0
1 When k = − , the equation is 3 1 y = − ( x − 1) + 4 . 3 ∴ x + 3 y − 13 = 0
3 . 5
158
Chapter 10 Straight Lines and Rectilinear Figures
18. (a) An equation of L is y − 3 = m( x + 5) , where m is real. (b) Rewrite L as mx − y + (5m + 3) = 0 mx y + =1 −5m + 3 5m + 3 5m + 3 x-intercept = − m y-intercept = 5m + 3 Area =
1 5m + 3 (− )(5m + 3) = 2 2 m
25m 2 + 30 m + 9 = ±4 m 25m 2 + 34 m + 9 = 0 or 25m 2 + 26 m + 9 ( m + 1)(25m + 9) = 0 or no solution 9 m = −1 or m = − 25 ∴ The equation of L 1 :
1 2 ∴ Slope of L 3 = 2
(b) Slope of PQ = −
Let the equation of L 3 be ( x − 2 y + 3) + k (5 x − 7 y + 6) = 0 (5k + 1) x − (7k + 2) y + (3 + 6k ) = 0 5k + 1 =2 Slope = 7k + 2 5k + 1 = 14 k + 4 1 k=− 3 ∴ The equation of L 3 is 2 x − y − 3 = 0 . (c) Substitute the coordinates of T into L 3 , 2(
4r − 2 3 )−( )−3= 0 1+ r 1+ r r=2
∴ The ration = 2 : 1
x + y + 2 = 0 or 9 x + 25 y − 30 = 0
Revision Exercise 10 (P.262) 2x − y + 8 = 0 19. (a) Solving , x − 4 y + 11 = 0 7 y − 14 = 0 y=2 x = −3 ∴ The point A is ( −3, 2) . (b) 2 x − y + 8 + k ( x − 4 y + 11) = 0 (2 + k ) x − (1 + 4 k ) y + (8 + 11k ) = 0 k+2 Slope of L = 4k + 1
1. Let (x, y) be the coordinates of P. −2 + 4r 2 + 6r x= , y= 1+ r 1+ r (Slope of OP)(Slope of AB) = −1
2 + 6r 6−2 ⋅ = −1 −2 + 4r 4 − ( −2) 4(6r + 2) = −6( 4r − 2) 1 r= 12 −1 + 2 −2 + 4 1 , ) = ( , 1) 2 2 2 Substitute into the line 1 a( ) + 3(1) − 5 = 0 2 a=4
2. Mid-point = (
(c) Slope of L 0 = 2
m−2 =1 1 + 2m 1 m = −3 or 3 1 ( x + 3) 3 y − 2 = −3 x − 9 or 3 y − 6 = x + 3 3x + y + 7 = 0 or x − 3 y + 9 = 0 y − 2 = −3( x + 3) or y − 2 =
3. Slope of the lines are
k=4 Distance =
20. (a) Let (x, y) be the coordinates of T.
8 − ( −2) 32 + 4 2
PT : TQ = r : 1 x=
4r − 2 3 , y= 1+ r 1+ r
∴ The point T is (
4r − 2 3 , ). 1+ r 1+ r
4. Equation of PM is 3 y + 2 = ( x − 1) 2 2 y + 4 = 3x − 3 3x − 2 y − 7 = 0
−3 −3 and . 4 k
=2
Chapter 10 Straight Lines and Rectilinear Figures
Equation of PN is
7. Equation of AB is 3 y + 2 = ( x + 1) 4 4 y + 8 = 3x + 3 3x − 4 y − 5 = 0
y − 1 = −2( x + 4) 2x + y + 7 = 0 3 x − 2 y − 7 = 0 2 x + y + 7 = 0
3 x − 4 y − 5 = 0 x − 2 y − 1 = 0
7x + 7 = 0 x = −1 y = −5
∴ The point P is ( −1, − 5) . 5. Let the ratio be 1 : r , the intersection of x − y + 1 = 0 and AB be P(a, b). 3r − 1 r+3 ∴ a= , b= 1+ r 1+ r Substitute (a, b) into x − y + 1 = 0 , 3r − 1 r + 3 ∴ − +1 = 0 1+ r 1+ r r =1 ∴ The required ratio is 1 : 1 . y
6.
L2: x − y + c = 0
A O
2y − 2 = 0 y =1 x=3 B = (3, 1)
AB = (3 + 1)2 + (1 + 2)2 = 5 8. Let (x, y) be the coordinates of P. ( x + 1)2 + ( y + 2) 2 = ( x − 1)2 + ( y − 4)2 4 x + 12 y − 12 = 0
4 x + 12 y − 12 = 0 4 x + 3 y − 12 = 0 9y = 0 y=0 x=3 ∴ The point P is (3, 0) . 9 − 10. No solutions are provided for the H.K.C.E.E. questions because of the copyright reasons.
B
C
159
L3: y = 1 x
L1: x + y = 2
Solve for the coordinates of A, x + y = 2 y =1 ∴ The coordinates of A are (1, 1). Solve for the coordinates of B, x+y=2 x − y + c = 0 c c ∴ The coordinates of B are (1 − , 1 + ) . 2 2 Solve for the coordinates of C, y =1 x − y + c = 0 ∴ The coordinates of C are (1 − c, 1). 1 c Area of ∆ABC = (1 − 1 + c)( ) 2 2 c2 = 4 =4 ∴ c 2 = 16 c = ±4
11. (a) Slope of the perpendicular =
−8 + 3 = −5 3−2
1 5 Equation of BC: 1 y + 8 = ( x − 3) 5 x − 3 = 5 y + 40 x − 5 y − 43 = 0 Slope of BC =
(b) Substitute B(−2, −a) into equation of BC, a = 9 , ∴ B is (−2, −9).
m AB =
−3 − ( −9) 3 = 2 − ( −2) 2
Equation of AC: 2 y + 3 = − ( x − 2) 3 2 x + 3y + 5 = 0 x − 5 y − 43 = 0 2 x + 3 y + 5 = 0 13 y = −91 y = −7 x=8 ∴ The point C is (8, − 7) .
160
Chapter 10 Straight Lines and Rectilinear Figures
3x + y − 2 = 0 12. (a) 5 x − 3 y − 8 = 0 14 x − 14 = 0 x =1 y = −1 ∴ The point P is (1, − 1) . Let R be (x, y), then x +1 y −1 =4, = −7 2 2 x=7, y = −13 ∴ R is (7, − 13) . (b) Let QR be 5 x − 3 y + k1 = 0 , RS be 3 x + y + k2 = 0 . Substitute R(7, −13) into QR, k1 = −74 . ∴ QR is 5 x − 3 y − 74 = 0 . Substitute R(7, −13) into RS, k2 = −8 . ∴ RS is 3 x + y − 8 = 0 . 13 − 16. No solutions are provided for the H.K.C.E.E. questions because of the copyright reasons. 17. Substitute (−1, 2) into the given equation, ∴
(2 − a)( −1) + a(2) + a 2 − 2 a = 0
∴
( a − 1)( a + 2) = 0
a = 1 or −2 Substitute a = 1 into the equation, x + y +1− 2 = 0 x + y −1 = 0
Substitute a = −2 into the equation, 4x − 2y + 4 + 4 = 0 2x − y + 4 = 0 18 − 20. No solutions are provided for the H.K.C.E.E. questions because of the copyright reasons. 21. (a) Slope of L 1 = 1 , slope of L 2 = −1 y − 3 = −1( x + 2) y − 3 = −x − 2 ∴ The equation of L 2 is x + y − 1 = 0 .
x−y=0 (b) x + y − 1 = 0 2y − 1 = 0 1 y= 2 1 x= 2 1 1 ∴ The point of intersection is ( , ) . 2 2
(c) Let (x, y) be the coordinates of the required point. Using the mid-point formula,
x−2 1 = 2 2 x=3 y+3 1 = 2 2 y = −2 ∴ The point is (3, − 2) . 22. (a) An equation of the family is 3 x − 4 y − c = 0 , where c is real. (b) Substitute (2, 3) into 3 x − 4 y − c = 0 , ∴
3(2) − 4(3) − c = 0 c = −6 ∴ The required equation is 3 x − 4 y + 6 = 0 .
(c) The distance = 6 − ( −15) 32 + 4 2 21 = 5 21 = 5
8+7
23. (a) d =
32 + 4 2
=3
(b) (i) Let θ be the acute angle, 3 1 ∴ sin θ = = 3 2 2 θ = 45° L1 P
L2 3 2
Q
θ
3
3 (ii) Slope of L 2 is − . Let the slope of L 4 be m. ∴
tan 45° =
m + 43
1 − 34m 3 3m m + = ±(1 − ) 4 4
m=
1 or m = −7 7
Chapter 10 Straight Lines and Rectilinear Figures
1 , the equation of L is 7 1 y − 3 = ( x − 2) 7 x − 7 y + 19 = 0
(c) When m =
When m = −7 , the equation of L is y − 3 = − 7( x − 2) 7 x + y − 17 = 0 24. No solution is provided for the H.K.C.E.E. question because of the copyright reasons.
Enrichment 10 (p.266) 1. (a) Equation of AB is −6 − 2 y−2= ( x + 4) 4+4 x+y+2=0 Substitute P(a, b) into the equation of AB, a+b+ 2=0 b = −a − 2
(b)
4 1 a 2 5 4
−6 −a−2 5 −6
= 12
4( − a − 2) + 5a − 30 + 6 a + 5( a + 2) − 20 = 24 12 a − 48 = 24 12 a − 48 = 24 or 12 a − 48 = −24 a = 6 or a = 2 2. y − 3 = m( x + 2) y = m( x + 2 ) + 3 Substitute into the curve,
1 ( x + 1)2 2 2 mx + 4 m + 6 = x 2 + 2 x + 1 m( x + 2 ) + 3 =
x 2 + 2(1 − m) x − 4 m − 5 = 0 Let ( x1, y1 ) be the coordinates of P, ( x2 , y2 ) be the coordinates of Q. x1 + x2 = 2( m − 1) Since C is the mid-point of PQ, x1 + x2 = −2 2 2( m − 1) = −4 2 m = −2 m = −1 The required equation is y − 3 = −( x + 2 ) x + y −1 = 0
161
3. (a) Let (x, y) be the coordinates of P. 5k − 2 x= 1+ k 5k + 3 y= 1+ k 5k − 2 5k + 3 , ) ∴ The point P = ( 1+ k 1+ k (b) Equation of AB is 5−3 y−3= ( x + 2) 5+2 2 x − 7 y + 25 = 0 2 x − 7 y + 25 = 0 x + y − 5 = 0
9 y − 35 = 0 35 y= 9 10 x= 9 10 35 , ) 9 9 5k − 2 10 = By (a), 1+ k 9 4 k= 5 ∴ AH : HB = 4 : 5 The point H = (
2x − y = 0 4. (a) 3 x + y − 10 = 0
5 x − 10 = 0 x=2 y=4 The point A is (2, 4). Q H is (2, −1). ∴ Equation of AD is x − 2 = 0 . (b) BE passes through H and perpendicular to AC. The equation of BE is 1 y + 1 = ( x − 2) 3 3y + 3 = x − 2 x − 3y − 5 = 0 x − 3y − 5 = 0 2 x − y = 0
5x + 5 = 0 x = −1 y = −2 The point B is ( −1, − 2) .
162
Chapter 10 Straight Lines and Rectilinear Figures
The point C is ( 4, − 2) .
Area of ∆ABC 2 2 2 2m 1 2m − 1 2m − 1 = 2 2m 2 2−m 2−m 2 2 1 2m 2 2m 2 2( = )+( )( ) + 2( ) 2 2m − 1 2m − 1 2 − m 2−m 2 2m 2 2m − 2( )−( )( ) − 2( ) 2m − 1 2m − 1 2 − m 2−m 1 4m 4 4 4m = + − − 2 2m − 1 2 − m 2m − 1 2 − m
Equation of BC is
=
−2 + 2 ( x + 1) 4 +1 y+2=0
1 8m − 4 m 2 + 8m − 4 − 8 + 4 m − 8m 2 + 4 m 2 (2 m − 1)(2 − m)
=
1 −12( m 2 − 2 m + 1) 2 (2 m − 1)(2 − m)
(c) CF passes through H and perpendicular to AB. The equation of CF is 1 y + 1 = − ( x − 2) 2 2y + 2 = −x + 2 x + 2y = 0
x + 2 y = 0 3 x + y − 10 = 0 5 x − 20 = 0 x=4 y = −2
y+2=
=
x − 2y + 2 = 0 5. (a) 2 x−y−2=0
6( m − 1)2 ( m − 2)(2 m − 1)
Classwork 1 (p.225)
3y − 6 = 0 y=2 x=2
1. AB = ( −2 − 4)2 + ( −1 − 1)2 = 40 = 2 10 AC = ( −2 − 2)2 + ( −1 + 5)2 = 4 2
The point A is (2, 2) .
BC = ( 4 − 2)2 + (1 + 5)2 = 40 = 2 10
(b) Equation of BC is y−0 =m x−0 y = mx
x − 2y + 2 = 0 (c) y = mx
Q AB = BC = 2 10 ∴ ABC is an isosceles triangle. 2. (a) AB = (0 − 2)2 + (2 − 2)2 = 2
BC = (2 − 2)2 + (2 − 8)2 = 6
x − 2 mx + 2 = 0 2 2m − 1 2m y= 2m − 1 2 2m The point B = ( , ). 2m − 1 2m − 1 x=
2 x − y − 2 = 0 y = mx
2 x − mx − 2 = 0 2 2−m 2m y= 2−m 2 2m The point C = ( , ). 2−m 2−m x=
AC = (0 − 2)2 + (2 − 8)2 = 40 = 2 10 (b) AB2 + BC 2 = 2 2 + 6 2 = AC 2 By the converse of Pyth. thm, ∆ABC is a right-angled triangle.
3.
(10 − p)2 + ( − p − 7)2 = 13 100 − 20 p + p 2 + p 2 + 14 p + 49 = 169
2 p 2 − 6 p − 20 = 0 p 2 − 3 p − 10 = 0 ( p − 5)( p + 2) = 0 p = 5 or p = −2
Chapter 10 Straight Lines and Rectilinear Figures
Classwork 2 (p.229) 3r =2 1+ r r=2 2 + 5r (ii) = −4 1+ r 2 r=− 3 (b) (i) Let (x, y) be the coordinates of P. −8 − 4(3) x= = −5 1+ 3 −4 + 8(3) =5 y= 1+ 3 ∴ The point P is ( −5, 5) . (ii) Let (x, y) be the coordinates of P. −1 + 8( 12 ) x= =2 1 + 12
1. (a) (i)
y=
4 − 8( 12 ) 1+
1 2
=0
∴ The point P is (2, 0) . (iii) Let (x, y) be the coordinates of P. −5 + 1( −2) x= =7 1− 2 1 + 5( −2) y= =9 1− 2 ∴ The point P is (7, 9) .
2. (a) Let (0, y) be the coordinates of C and
AC : CB = r : 1 . r−3 =0 r +1 r=3 ∴
AC : CB = 3 : 1
3( −6) + 2 = −4 3 +1 ∴ The point C is (0, − 4) .
1 [ 46 − ( −26)] square units 2 = 36 square units (b) Area −2 5 −6 −1 1 4 −5 square units = 2 2 2 −2 5 1 = {[( −2)( −1) + ( −6)( −5) + ( 4)(2) + (2)(5)] 2 − [( −6)(5) + ( 4)( −1) + (2)( −5) + ( −2)(2)]} square units 1 = [50 − ( −48)] square units 2 = 49 square units =
2. (a) Area of ∆ABC
1 = 2 =
−4 k 2 3 5 6+k −4 k
square units
1 [( −4)(3) + (2)(6 + k ) + (5)( k )] − [(2)( k ) 2 + (3)(5) + (6 + k )( −4)] square units
1 7k + 9 + 2 k square units 2 9 = k + 1 square units 2
=
(b) If area of ∆ABC = 18 square units, 9 k + 1 = 18 2 k +1 = 4 k +1 = 4 or k = 3 or k = −5
k + 1 = −4
(b) y =
Classwork 4 (p.236) 1. (a) m = tan α =
Classwork 3 (p.233) 1. (a) Area
5 −3 1 −5 8 = square units 2 −7 3 5 −3 1 = {[(5)(8) + ( −5)(3) + ( −7)( −3)] 2 − [( −5)( −3) + ( −7)(8) + (5)(3)]} square units
163
3 − ( −5) 8 =− 1− 4 3
α = 111° (corr. to 3 sig. fig.)
(b) m = tan α =
1− 2 1 =− 4 − ( −1) 5
a = 169° (corr. to 3 sig. fig.) (c) m = tan α = α = 0°
−2 − ( −2) =0 0 − ( −3)
164
Chapter 10 Straight Lines and Rectilinear Figures
5 − ( −6) 2−2 ∴ m is undefined.
2. (a) Let θ be the required angle.
(d) m = tan α =
tan θ =
α = 90°
2 3 1 + 2( 23 )
=
4 7
θ = 29.7° (corr. to 3 sig. fig.) (b) Let θ be the required angle.
5−k =1 2−0 k=3 0 − ( −2) (b) =3 k−5
4 − ( −1) 5 = 1 + 4( −1) 3 θ = 59.0° (corr. to 3 sig. fig.)
2. (a)
k=5
2−
tan θ =
2 3
Classwork 5 (p.239) 1. Slope of AB =
−2 − ( −6) = −2 −4 − ( −2)
Slope of BC =
6 − ( −2) = −2 −8 − ( −4)
∴ AB // BC They are parallel and have a common point B. Therefore A, B, C are collinear.
7 − ( −6) 13 = 7 − ( −2) 9 −3 − 6 −9 Slope of CD = = 7 − ( −6) 13 13 −9 Product of slopes = ( )( ) = −1 9 13 ∴ AB ⊥ CD
Classwork 7 (p.243) 1. (a) y = − x + b 5 = −1 + b b=6
5 x+b 3 5 3 = (3) + b 3 b = −2
(b) y =
2.
Equation of straight line Point Slope y-intercept
Point-slope form
2. Slope of AB =
Classwork 6 (p.240) 1. Let C(0, b) and D(a, 0) be the intersections of AB with the coordinate axes. Then slope of AC = slope of AB b − ( −3) 9 − ( −3) = 0 − ( −4) 4 − ( −4) 8b + 24 = 48 b=3 ∴ y-intercept is 3 . Slope of AD = Slope of AB 0 − ( −3) 9 − ( −3) = a − ( −4) 4 − ( −4) 24 = 12 a + 48 a = −2 ∴ x-intercept is −2 .
Slopeintercept form
(−2, 6)
−1
4
(y − 6) = −(x + 2) y = −x + 4
(1, 4)
2
2
(y − 4) = 2(x − 1) y = 2x + 2
(−2, 0) −
5 2
−5
y=−
4
−3
y = 4x − 3
−2
5
y = −2x + 5
5 (x + 2) 2
y=−
5 x−5 2
Classwork 8 (p.246) 1. (a) By using the two-point form, the equation of the line is 6−3 y−3=( )( x − 2) −1 − 2 y − 3 = −( x − 2 ) x+y−5=0 (b) By using the two-point form, the equation of the line is 1+ 2 y+2=( )( x + 3) 5+3 3 y + 2 = ( x + 3) 8 3x − 8y − 7 = 0
Chapter 10 Straight Lines and Rectilinear Figures
(c) y = 2
2 ∴ The slope is − . 3 y-intercept is 0 .
(d) x = 3 (e) y = 0
Substitute y = 0 into the equation,
(f) By using the intercept form, the equation of the line is x y + =1 3 4 4 x + 3 y − 12 = 0 2. (a) x + 2 y − 3 = 0 2y = −x + 3 1 3 y=− x+ 2 2 1 ∴ The slope is − . 2 3 y-intercept is . 2 Substitute y = 0 into the equation, ∴ x−3= 0 x=3 ∴ x-intercept is 3 . (b) 3 y − x + 1 = 0 3y = x − 1 1 1 y= x− 3 3 1 ∴ The slope is . 3 1 y-intercept is − . 3 Substitute y = 0 into the equation,
∴
x=0 x-intercept is 0 . (e) 2 − x = 0 x=2 ∴ The slope is undefined.
There is no y-intercept. x-intercept is 2 . (f) y + 5 = 0 y = −5 ∴ The slope is 0 . y-intercept is −5 . There is no x-intercept.
Classwork 9 (p.249)
L1 1
O
(c) 2 y = 1 − 4 x
1 2 ∴ The slope is −2 .
1 . 2
Substitute y = 0 into the equation, ∴
0 = 1 − 4x 1 x= 4
1 ∴ x-intercept is . 4 (d) 3 y + 2 x = 0 3 y = −2 x 2 y=− x 3
x
π 6
In the figure, the equation of L1 is π π x cos + y sin − 1 = 0 6 6 3 1 x + y −1 = 0 2 2 3x + y − 2 = 0
−x + 1 = 0 x =1 ∴ x-intercept is 1 .
y = −2 x +
y
(a)
∴
y-intercept is
165
y
(b)
L2 3
2π 3
O
In the figure, the equation of L 2 is 2π 2π x cos + y sin −3= 0 3 3 1 3 − x+ y−3= 0 2 2 x − 3y + 6 = 0
x
166
Chapter 10 Straight Lines and Rectilinear Figures
y
(c)
∴
p=
L3
2 2 , sin θ = 2 2 ∴ θ = 45° (c) The normal form is 3x − 2 y + 7 cos θ =
x
O
In the figure, the equation of L 3 is π π x cos + y sin = 0 4 4 2 2 x( ) + y( )=0 2 2 x+y=0
4
O
x
In the figure, the equation of L 4 is x cos 0 + y sin 0 − 4 = 0 x−4=0
Classwork 10 (p.251) (a) The normal form is
3x + 4 y − 5
=0
32 + 4 2 3 4 x + y −1 = 0 5 5 p =1
3 4 , sin θ = 5 5 θ = 53.1° (corr. to 3 sig. fig.)
cos θ = ∴
∴
7 13
p=
3 2 , sin θ = 13 13 ∴ θ = 146° (corr. to 3 sig. fig.) (d) The normal form is x − 3y =0 − 12 + 32 1 3 − x+ y=0 10 10
L4
∴
=0 − 32 + 2 2 3 2 7 − x+ y− =0 13 13 13
cos θ = −
y
(d)
2 2
(b) The normal form is
∴
p=0
1 3 , sin θ = 10 10 ∴ θ = 108° (corr. to 3 sig. fig.) (e) The normal form is y+5 =0 − 12 −y − 5 = 0 cos θ = −
∴ p=5 cos θ = 0 , sin θ = −1 ∴ θ = 270° (f) The normal form is x−3 =0 12 x−3= 0 ∴ p=3 cos θ = 1 , sin θ = 0 ∴ θ = 0°
−x − y + 1
=0 − 12 + 12 1 1 1 x+ y− =0 2 2 2 2 2 2 x+ y− =0 2 2 2
Classwork 11 (p.252) 1. (a) By considering the distance from (4, −2) to the line, d=
2( 4) − ( −2) + 7 2 2 + ( −1)2
=
17 5 units 5
Chapter 10 Straight Lines and Rectilinear Figures
(b) By considering the distance from (−2, 3) to the line,
d=
−4( −2) + 3(3) − 1 ( −4)2 + 32
16 units = 5
2. Let (s, 0) be the coordinates of P.
5s + 36
=2 52 + ( −12)2 5s + 36 =2 13 5s + 36 5s + 36 or =2 = −2 13 13 5s = −10 or 5s = −62 62 s = −2 or s=− 5 62 ∴ The point P is ( −2, 0) or ( − , 0) . 5 3. Let P(x, y) be a point on the perpendicular bisector, then the distance from P to the two lines are equal. 2x − 4y + 1
=
or 2x − 4y + 1 = 4x + 2y − 6 2 x + 6 y − 7 = 0 or
2x − 4y + 1 = −( 4 x + 2 y − 6) 6x − 2y − 5 = 0
Classwork 12 (p.254) 1. The distance between L1 and L 2 is d=
8 − ( −2) 4 2 + 32
Classwork 13 (p.255) (a) An equation of L is 3 x − y + k = 0 , where k is real. (b) Mid-point of AB −1 + 3 2 + 3 5 =( , ) = (1, ) 2 2 2 Substitute the point into L, 5 3(1) − + k = 0 2 1 k=− 2 ∴ An equation of L1 is 6 x − 2 y − 1 = 0 . (c) Let the equation of L 2 be 3 x − y + k2 = 0 . 3( −1) − 2 + k2
= 2 10 32 + 12 k2 − 5 = ±2 10 10 k2 − 5 = 2(10) or k2 − 5 = −2(10) or k2 = 25 k2 = −15 ∴ The equation of L is 3 x − y + 25 = 0 or 3 x − y − 15 = 0 .
2x + y − 3
22 + 42 2 2 + 12 2x − 4y + 1 2x + y − 3 = 2 5 5 2 x − 4 y + 1 = ±( 4 x + 2 y − 6)
units
= 2 units 2. Let the equation of L' be x + y − p = 0 . −1 − ( − p)
= 2 12 + 12 −1 + p =± 2 2 −1 + p = 2 or −1 + p = −2 p = 3 or p = −1
∴ The equation of L' is x + y − 3 = 0 or x + y + 1 = 0 .
167
Classwork 14 (p.256) (a) An equation of L is y − 3 = k[ x − ( −5)] y = k ( x + 5) + 3 , where k is real (b) Since L1 passes through (2, 1), 1 = k (2 + 5) + 3 2 k=− 7 ∴ The equation of L1 is 2 x + 7 y − 11 = 0 . (c) From the equation of L y = k ( x + 5) + 3 y − kx = 5k + 3 − kx y + =1 5k + 3 5k + 3 x-intercept 5k + 3 = −1 k k = 5k + 3 3 k =− 4 ∴ The equation of L 2 is 3 y = − ( x + 5) + 3 4 3x + 4 y + 3 = 0 =−
168
Chapter 10 Straight Lines and Rectilinear Figures
Classwork 15 (p.258) (a) An equation of L is 3 x − 2 y − 1 + k ( x + y + 3) = 0 , where k is real. (b) (i) Rearrange the equation as (3 + k ) x + ( k − 2) y + (3k − 1) = 0 1 Since the slope is , 4 3+ k 1 − = k−2 4 k − 2 = −12 − 4 k 5k = −10 k = −2 Substitute k = −2 into (1), (3 x − 2 y − 1) − 2( x + y + 3) = 0 x − 4y − 7 = 0 (ii) Since y-intercept is 2, −
3k − 1 =2 k−2 2 k − 4 = −3k + 1 5k = 5 k =1
Substitute k = 1 into (1), 3x − 2 y − 1 + x + y + 3 = 0 4x − y + 2 = 0