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Active Maths 2 (Strands 1–5): Ch 20 Solutions
Chapter 20 Exercise 20.1
1
Q. 1. (i)
x
y
A = (2,2)
B = (–4,6)
–6
–4
–2
2
4
6
–2 2 4 6 8 10 12 1400
–4–6
(ii) C ( 2 − 4 ______ 2 , 2 + 6 ______ 2 ) = (−1,4)
(iii) |AC| = √___________________
(−1 − 2)2 + (4 − 2)2
= √_____________
(−3)2 + (−2)2
= √______
9 + 4
= √___
13
|CB| = √___________________
(−1 + 4)2 + (4 − 6)2
= √___________
(3)2 + (−2)2
= √______
9 + 4
= √___
13
∴ |AC| = |CB|
(iv) Slope of AB = 6 − 2 _______ −4 − 2 = 4 ___ −6 = –2 ___ 3
y − 2 = – 2 __ 3 (x − 2)
3y − 6 = –2(x − 2) 3y – 6 = –2x + 4 2x + 3y – 10 = 0
(v) AB cuts the y-axis at x = 0. 0 + 3y – 10 = 0 3y = 10
y = 10 ___ 3
|AB| cuts the y-axis at ( 0, 10 ___ 3 ) .
(vi) AB cuts the x-axis at y = 0
2x + 3(0) – 10 = 0
2x = 10
x = 5
AB cuts the x-axis at (5,0).
Q. 2. (i)
x
y
A = (2,4)
C = (–2,0) B = (6,0)
–6
–4
–2
2
4
6
–2 2 4 6 8 10 12 1400
–4–6
(ii) M ( 6 − 2 ______ 2 , 0 + 0 ______ 2 ) = M(2,0)
(iii) AB: slope = 0 − 4 ______ 6 − 2 = −4 ___ 4 = −1
(iv) y − y1 = m(x − x1)
y − 4 = −1(x − 2)
y − 4 = −x + 2
x + y − 6 = 0
(v) x + y = 6 x – y = –2
2x = 4
x = 2
2 + y = 6
y = 4
(2,4)
(vi) h = 4 base = 8
Area = 1 __ 2 × b × ⊥h
= 1 __ 2 × 8 × 4
= 16 units2x
y
–4–3
–2–1
123456
0–1 1 2 3 4 5 6 7 8 9 10 11 120–2–3–4–5–6–7
C = (–2,0)
A = (2,4)
B = (6,0)
2 Active Maths 2 (Strands 1–5): Ch 20 Solutions
Q. 3. (i) m: x + 4y − 8 = 0
4y = −x + 8
y = −1 ___ 4 x + 2
Slope = −1 ___ 4 , C = 2
x
y
–5–4–3–2–1
123456
00–1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17–2–3–4
n
m
n: x − y – 3 = 0
−y = − x + 3
y = x − 3
Slope = 1, C = −3
(ii)
x
y
–5–4–3–2–1
123456
00–1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17–2–3–4
n
m(4,1)
(4,1)
(iii) x + 4y – 8 = 0 – x –+ y –+ 3 = 0
5y − 5 = 0
5y = 5
y = 1
x − 1 − 3 = 0
x − 4 = 0
x = 4
(4,1)
(iv) At y-axis, x = 0
0 + 4y − 8 = 0
4y = 8
y = 2
(0,2)
3Active Maths 2 (Strands 1–5): Ch 20 Solutions
Q. 4. (i) 2x − y + 3 = 0
−y = −2x − 3
y = 2x + 3
Slope = 2, C = 3
5x + y − 10 = 0
y = −5x + 10
Slope = −5, C = 10
x0
0–1–1
123456789
10
1 2 3 4 5 6 7 8 9 10 11 12–2–3–4–5–6–7
s ry
(ii)
x0
0–1–1
123456789
10
1 2 3 4 5 6 7 8 9 10 11 12–2–3–4–5–6–7
s ry
(1,5)
(1,5)
(iii) 5x + y – 10 = 0 2x – y + 3 = 0
7x − 7 = 0
7x = 7
x = 1
5(1) + y − 10 = 0
y − 5 = 0
y = 5
(1,5)
(iv) At x-axis, y = 0
2x − 0 + 3 = 0
2x = − 3
x = −3 ___ 2
( −3 ___ 2 ,0 )
4 Active Maths 2 (Strands 1–5): Ch 20 Solutions
Q. 5. (i) 3x + 2y = 18
2y = − 3x + 18
y = −3 ___ 2 x + 9
Slope = −3 ___ 2 , c = 9
y = 3x
Slope = 3, c = 0
–6 –4 –2
2
46
810
0 2 4 6 8 10 12 14 16 18x
yp q
(ii)
–6 –4 –2
2
46
810
0 2 4 6 8 10 12 14 16 18x
yp q
(2,6)
(2,6)
(iii) 3x + 2y = 18 3x + 2(3x) = 18 3x + 6x = 18 9x = 18 x = 2
y = 3x
y = 3(2) y = 6
(2,6)
Q. 6. (i) d = √_________________
(5 + 3)2 + (4 + 2)2
= √_______
82 + 62
= √____
100
= 10 units
(ii) 10 × 1 ___ 10 = 1 km
(iii) T = D __ S = 1 ___ 2.5 = 2 __ 5 hr = 24 minutes
Q. 7. (i) h: 3x + y = 5
(1,2) ⇒ 3(1) + 2 = 5?
3 + 2 = 5?
5 = 5
∴ (1,2) ∈ h
(ii) (k,0) ⇒ 3(k) + 0 = 5
3k = 5
k = 5 __ 3
(iii) Let y = 3
⇒ 3x + 3 = 5
3x = 2
x = 2 __ 3
( 2 __ 3 ,3 ) Let y = −1
⇒ 3x − 1 = 5
3x = 6
x = 2
(2,−1)
Let y = 5
⇒ 3x + 5 = 5
3x = 0
x = 0
(0,5)
(iv) Slope = 5 + 1 ______ 0 − 2 = 6 ___ −2 = −3
(v) 3x + y = 5
y = −3x + 5
m = −3
Q. 8. (i) y = 4x − 3
(ii) y = −3x + 5
Q. 9. (2,4) → (3,−1) → (4,−6)
a = 4
b = −6
Q. 10. (i) m = 2 − 6 _______ 12 − 4 = −4 ___ 8 = −1 ___ 2
y − y1 = m(x − x1)
y − 6 = −1 ___ 2 (x − 4)
y − 6 = −1 ___ 2 x + 2
2y − 12 = − x + 4
l1: x + 2y = 16
5Active Maths 2 (Strands 1–5): Ch 20 Solutions
(ii) l2: y = −4x + 0
4x + y = 0
(iii) x + 2y = 16
x + 2(−4x) = 16
x − 8x = 16
−7x = 16
x = −16 ____ 7
y = −4 ( −16 ____ 7 ) y = 64 ___ 7
( −16 ____ 7 , 64 ___ 7 ) Midpoint ( 4 − 16 __ 7
______ 2 , 6 + 64 __ 7
______ 2 ) = ( 12 ___ 14 , 108 ____ 14 ) = ( 6 __ 7 , 54 ___ 7 ) Q. 11. (a) A(1,1)
B(11,1)
(b)
20181614121086420–2
2
46
8
–2
0–4
x
y D = (6,8)
B = (11,1)A = (1,1)
(c)
20181614121086420–2
2
46
8
–2
0–4
x
y D = (6,8)
B = (11,1)A = (1,1) C = (6,1)
(6,1)
(d)
20181614121086420–2
2
46
8
–2
0–4
x
y D = (6,8)
B = (11,1)A = (1,1) C = (6,1)
(6,1)
(e) |AD| = √_________________
(6 − 1)2 + (8 − 1)2
= √_________
(5)2 + (7)2
= √________
25 + 49
= √___
74
|BD| = √__________________
(6 − 11)2 + (8 − 1)2
= √___________
(−5)2 + (7)2
= √________
25 + 49
= √___
74
(f) ABD is an isosceles Δ as two sides are equal in measure.
(g) |AD| = |BD| shown in (e)
|AC| = |BC| as C is the midpoint
CD| = |CD| as it is the common side
∴ Δ ACD ≡ Δ BCD by SSS.
Q. 12. (i)
D = (4,8)
A = (2,3)
B = (10,4)
x
y
C = (12,9)
6 Active Maths 2 (Strands 1–5): Ch 20 Solutions
(ii) |AD| = √_________________
(4 − 2)2 + (8 − 3)2
= √_________
(2)2 + (5)2
= √_______
4 + 25
= √___
29
|BC| = √___________________
(12 − 10)2 + (9 − 4)2
= √_________
(2)2 + (5)2
= √_______
4 + 25
= √___
29
∴ |AD| = |BC|
(iii) Midpoint of [AC]:
E ( 2 + 12 _______ 2 , 3 + 9 ______ 2 ) = ( 14 ___ 2 , 12 ___ 2 ) = (7,6)
Midpoint of [BD]
F ( 10 + 4 _______ 2 , 4 + 8 ______ 2 ) = ( 14 ___ 2 , 12 ___ 2 ) = (7,6)
E = F
∴ [AC] and [BD] bisect each other.
(iv) Yes, I can now conclude that ABCD is a parallelogram as opposite sides are equal in length and the diagonals bisect each other.
Exercise 20.2
Q. 1. Slope Slope of parallel line
Slope of perpendicular
line– 1 _ 2 – 1 _ 2 2
3 _ 4 3 _ 4 – 4 _ 3
– 1 _ 3 – 1 _ 3 3
– 8 __ 11 – 8 __ 11 11 __ 8
13 13 – 1 __ 13
Q. 2. 5 __ 4 Q. 3. 4 __ 3 Q. 4. − 1 __ 2
Q. 5. −4 Q. 6. 1 __ 3
Q. 7. (i) 2x + y − 8 = 0
y = −2x + 8
m = −2
(ii) 3x − y + 6 = 0
− y = −3x − 6
y = 3x + 6
m = 3
(iii) 2x − 2y = 9
−2y = −2x + 9
y = x − 9 __ 2
m = 1
(iv) 3(x − y) = 4
x − y = 4 __ 3
−y = −x + 4 __ 3
y = x − 4 __ 3
m = 1
(v) 2x − 3(y − 2) = 0
2x − 3y + 6 = 0
−3y = −2x − 6
y = 2 __ 3 x + 2
m = 2 __ 3
(vi) 4x = 6 − 5y
5y = −4x + 6
y = −4 ___ 5 x + 6 __ 5
m = − 4 __ 5
Q. 8. 3x + y = 6
y = − 3x + 6 m = −3 c = 6 Line ⊥ 3x + y = 6 has m = 1 __ 3
∴ Equation is y = 1 __ 3 x + 6
3y = x + 18
Q. 9. 2x − y = 3 −y = −2x + 3 y = 2x −3 m = 2, c = −3 Line ⊥ to 2x − y = 3 has slope − 1 __ 2
∴ y = −1 ___ 2 x − 3
2y = −x −6
x + 2y + 6 = 0
7Active Maths 2 (Strands 1–5): Ch 20 Solutions
Q. 10. x + 2y = 4 2y = −x + 4
y = −1 ___ 2 x + 2
m = −1 ___ 2 , c = 2
Slope of line ⊥ to x + 2y = 4 is 2
∴ y = 2x + 2
Q. 11. x − 3y + 6 = 0
−3y = −x −6
y = 1 __ 3 x + 2
m = 1 __ 3 , c = 2
Line || to x − 3y + 6 = 0 has slope 1 __ 3
y − y1 = m(x − x1)
y − 4 = 1 __ 3 (x − 2)
3y − 12 = x − 2
x − 3y + 10 = 0
Q. 12. 2x + 3y + 6 = 0
3y = −2x − 6
y = −2 ___ 3 x − 2
m = −2 ___ 3 , c = −2
Line ⊥ to 2x + 3y + 6 = 0 has slope 3 __ 2
y − y1 = m(x − x1)
y + 6 = 3 __ 2 (x − 1)
2y + 12 = 3x − 3
3x − 2y − 15 = 0
Q. 13. (i) a: 2x + 3y − 8 = 0
3y = −2x + 8
y = −2 ___ 3 x + 8 __ 3
b: 2x − 3y + 6 = 0
−3y = −2x − 6
y = 2 __ 3 x + 2
m1m2 = ( −2 ___ 3 ) ( 2 __ 3 ) ≠ 1
∴ a ⊥ b
(ii) a: −2x + 3y − 8 = 0
3y = 2x + 8
y = 2 __ 3 x + 8 __ 3
b: 2x − 3y + 14 = 0
−3y = −2x − 14
y = 2 __ 3 x + 14 ___ 3
m1 = m2
∴ a ⊥ b
(iii) a: 2x − 6y – 8 = 0
− 6y = −2x + 8
y = 1 __ 3 x − 8 __ 6
b: x − 3y + 1 = 0
−3y = −x −1
y = 1 __ 3 x + 1 __ 3
m1 = m2
∴ a ⊥ b
(iv) a: 2x − 4y − 3 = 0
−4y = −2x + 3
y = 1 __ 2 x − 3 __ 4
b: y = −2x + 10
m1m2 = ( 1 __ 2 ) (−2) = −1
∴ a ⊥ b
(v) a: −2x + 3y − 3 = 0
3y = 2x + 3
y = 2 __ 3 x + 1
b: 3y = −2x + 10
y = −2 ___ 3 x + 10 ___ 3
m1m2 = ( 2 __ 3 ) ( −2 ___ 3 ) ≠ −1
∴ a ⊥ b
(vi) a: y = 2x − 7
b: x +2y = 3
2y = −x + 3
y = −1 ___ 2 x + 3 __ 2
m1m2 = (2) ( −1 ___ 2 ) = −1
∴ a ⊥ b
Q. 14. (i) (1,2)
y = 3 is horizontal,
∴ Line ⊥ to y = 3 must be vertical.
Vertical line going through (1,2) has equation x = 1.
8 Active Maths 2 (Strands 1–5): Ch 20 Solutions
(ii) It is not possible to verify as slope of y = 3 is 0 and slope of x = 1 is undefined.
Q. 15. l: 5x − y + 4 = 0
−y = −5x − 4
y = 5x + 4
m = 5
Slope of k = 5
y − y1 = m(x − x1)
y + 3 = 5(x − 4)
y + 3 = 5x − 20
5x − y − 23 = 0
Q. 16. m: 3x − 6y + 11 = 0
−6y = − 3x − 11
y = 1 __ 2 x + 11 ___ 6
n has slope −2
y − y1 = m(x − x1)
y − 2 = −2(x + 4)
y − 2 = −2x − 8
2x + y + 6 = 0
Q. 17. p: 7x − y + 14 = 0
−y = −7x − 14
y = 7x + 14
q has slope 7
y − y1 = m(x − x1)
y −3 = 7(x + 1)
y − 3 = 7x + 7
7x − y + 10 = 0
Q. 18. l: x − y + 4 = 0
−y = − x − 4
y = x + 4
k has slope −1
y − y1 = m(x − x1)
y − 2 = −1(x − 2)
y − 2 = −x + 2
x + y − 4 = 0
Revision Exercises
Q. 1. (i) m = rise ___ run = 6 __ 5
(ii) C ( 1 + 6 ______ 2 , − 2 + 4 ________ 2 ) = ( 3 1 _ 2 ,1 ) (iii) |AB| = √
___________________ (1 − 6)2 + (− 2 − 4)2
= √______________
(− 5)2 + (− 6)2
= √________
25 + 36
= √___
61
(iv) Slope of k = −5 ___ 6
C ( 3 1 _ 2 ,1 ) y − y1 = m(x − x1)
y − 1 = −5 ___ 6 ( x − 7 __ 2 ) 6y − 6 = − 5x + 35 ___ 2
12y − 12 = –10x + 35
k: 10x + 12y − 47 = 0
(v) At x-axis, y = 0 10x − 47 = 0 10x = 47 x = 4.7 (4.7,0)
At y-axis, x = 0 12y − 47 = 0
12y = 47
y = 47 ___ 12 ( 0, 47 ___ 12 ) Q. 2. (i) l: 3x − 4y + 7 = 0
P(−1,h): 3(−1) − 4(h) + 7 = 0
−3 −4h + 7 = 0
− 4h + 4 = 0
− 4h = − 4
h = 1
m: 4x + 3y − 24 = 0
Q(k,0): 4(k) + 3(0) − 24 = 0
4k − 24 = 0
4k = 24
k = 6
9Active Maths 2 (Strands 1–5): Ch 20 Solutions
(ii) 3x − 4y + 7 = 0 × 3 4x + 3y − 24 = 0 × 4
9x − 12y + 21 = 0 16x + 12y − 96 = 0
25x − 75 = 0
25x = 75
x = 3
3x − 4y + 7 = 0 3(3) − 4y + 7 = 0 9 − 4y + 7 = 0 −4y = −16 y = 4
(3,4)
(iii)
x
m
l
R = (3,4)
P = (–1,1) Q = (6,0)–6 –4 –2
–2
2
468
–4
00
2 4 6 8 10 12 14 16 18
(iv) ∠PRQ is a right angle
⇒ PR ⊥ RQ
⇒ l ⊥ m
l: 3x − 4y + 7 = 0
−4y = −3x − 7
y = 3 __ 4 x + 7 __ 4
m: 4x + 3y − 24 = 0
3y = −4x + 24
y = −4 ___ 3 x + 8
m1m2 = ( 3 __ 4 ) ( −4 ___ 3 ) = −1
∴ l ⊥ m
∴ ∠PRQ is a right angle.
Q. 3. (i) P(−1,2) R(3,4)
M ( −1 + 3 _______ 2 , 2 + 4 ______ 2 ) = M ( 2 __ 2 , 6 __ 2 ) = M(1,3)
(ii) m = 4 − 2 ________ 3 − (−1) = 2 __ 4 = 1 __ 2
(iii) Line ⊥ to [PR] has slope −2 y − y1 = m(x − x1) y − 3 = − 2(x − 1) y − 3 = −2x + 2 2x + y − 5 = 0 (iv) k: x − 2y = 0 l: 2x + y = 5 × 2 x − 2y = 0 4x + 2y = 10 5x = 10 x = 2
x − 2y = 0 2 − 2y = 0 − 2y = − 2 y = 1 N(2,1)
Q. 4. l: 2x + 3y = − 1
(i) 3y = −2x −1
y = −2 ___ 3 x − 1 __ 3
Slope of k = 3 __ 2
y − y1 = m(x − x1)
y − 2 = 3 __ 2 (x − 3)
2y − 4 = 3x − 9
k: 3x − 2y −5 = 0
(ii) k and l are ⊥
2x + 3y = −1 ×2
3x − 2y = 5 ×3
4x + 6y = −2 9x − 6y = 15
13x = 13 x = 1
2x + 3y = −1 2(1) + 3y = −1 3y = −3 y = −1
Point of intersection of k and l is (1,−1)
P (3,2) → (1,−1) → P‘ (−1,−4)
10 Active Maths 2 (Strands 1–5): Ch 20 Solutions
Q. 5. (i) AD: m1 = −3 − (−1)
__________ 1 − 2 = −3 + 1 _______ −1
= −2 ___ −1 = 2
BC: m2 = 1 − (−2)
________ −2 − 4 = 1 + 2 ______ −6
= 3 ___ −6 = −1 ___ 2
m1m2 = 2 ( −1 ___ 2 ) = −1
∴ AD ⊥ BC
(ii) |BC| = √_____________________
(−2 − 4)2 + (1 − (−2))2
= √___________
(−6)2 + (3)2
= √_______
36 + 9
= √___
45
(iii) Area of ΔABC = 1 __ 2 × b × ⊥h
= 1 __ 2 × √___
45 × √__
5
= 1 __ 2 √___
90
= √__
9 √___
10 _______ 2 = 3 __ 2 √___
10 units2
Q. 6. (i) l: 3x −5y + 15 = 0
At x-axis, y = 0
3x + 15 = 0
3x = −15
x = −5
(−5,0) = C
m: 3x + 4y − 12 = 0
3x −12 = 0
3x = 12
x = 4
D = (4,0)
(ii) 3x − 5y + 15 = 0 – 3x
+– 4y –+ 12 = 0
−9y + 27 = 0
−9y = −27
y = 3
3x − 5(3) + 15 = 0
3x − 15 + 15 = 0
3x = 0
x = 0
E = (0,3)
(iii)
x
y
–210–8 –6 –4 –2–2
246
–4
00
2 4 6 8 10 12 14
E = (0,3)C = (–5,0) D = (4,0)
(iv) Area ΔCDE = 1 __ 2 × b × ⊥h
= 1 __ 2 × 9 × 3
= 27 ___ 2
= 13.5 units2
Q. 7. (i) R ( 2 − 2 ______ 2 , −3 + 1 _______ 2 ) = R ( 0 __ 2 , −2 ___ 2 ) = R(0,−1)
(ii) Slope PQ = 1 − (−3)
________ −2 − 2 = 1 + 3 ______ −4
= 4 ___ −4 = −1
∴ Slope of k = 1
k: y − y1 = m(x − x1)
y + 1 = 1 (x − 0)
y + 1 = x
k: y = x − 1
(iii) k: y = x − 1
S(3,2) ⇒ 2 = 3 − 1?
2 = 2
∴ S ∈ k
(iv) If ΔPQS is isosceles then |PS| = |QS|
|PS| = √___________________
(2 − 3)2 + (− 3 −2)2
= √_____________
(−1)2 + (−5)2
= √_______
1 + 25
= √___
26
|QS| = √__________________
(−2 −3)2 + (1 − 2)2
= √_____________
(−5)2 + (−1)2
= √_______
25 + 1
= √___
26
∴ ΔPQS is isosceles
11Active Maths 2 (Strands 1–5): Ch 20 Solutions
Q. 8. (a) Slope of l1 = rise ___ run = 275 ____ 25 = 11 ___ 1 = 11
Slope of l2 = rise ___ run = 340 ____ 40 = 17 ___ 2 = 8.5
(b) At a steady speed of 60 km/hr John’s car can travel 11 km per litre of fuel consumed and at a steady speed of 100 km/hr John’s car travels 8.5 km per litre of fuel consumed.
(c) At 60 km/hr: 200 ____ 11 × 149.9
= 2,725 5 __ 11 c
= €27.25
At 100 km/hr: 200 ____ 8.5 × 149.9
= 3,527 1 __ 17 c
= €35.27
€35.27 − €27.25 = €8.02 cheaper
Q. 9. (a) Line 4 has greatest slope as slope = 5
(b) Lines 1 and 2 are parallel as they both have slope 3.
(c)
x
y
–4 –3 –2 –1–1
123
–2–3–4–5–6–7–8
0 1 2 3 4 5 6 7 8 9 10 11
(d) It represents line 5.
(e) Line 6
(f) y = x − 7 – y = – 4x –+ 16
0 = −3x + 9
−9 = −3x
3 = x
y = 3 − 7
y = −4
(3,−4)
(g) Line 4: If x = 3, y = 3 − 7 = –4
Line 6: If x = 3 , y = 4(3) −16
= 12 −16
= −4
12 Active Maths 2 (Strands 1–5): Ch 20 Solutions
Q. 10. (a) A(1,4)
(b)
x
D = (0,4)A = (1,4)
B = (2,0)E = (–6,0)
F = (4,–4)C = (–4,4)
–6–8–10 –4 –2–2
2
46
–4
00
2 4 6 8 10 12 14
(c) Midpoint [DF]
= ( 0 + 4 ______ 2 , 4 − 4 ______ 2 ) = ( 4 __ 2 , 0 __ 2 ) = (2,0)
(d) Slope BF = rise ___ run = −2 ___ 1 = −2
y intercept BF = D = 4
∴ BF: y = −2x + 4
(e) Slope CE = rise ___ run = −2 ___ 1 = −2
(f) y − y1 = m(x − x1)
y − 0 = −2(x + 6)
y = −2x − 12
2x + y + 12 = 0
(g) Area ΔBCE : Area ΔBCF
1 : 1
(h) Yes, ΔBCE and ΔBCF are congruent because they have equal area, |BE| = |CF| and |EC| = |BF|
Q. 11. (i) k: 4x−3y + 12 = 0
At x-axis, y = 0
4x + 12 = 0
4x = −12
x = −3
A (−3,0)
At y-axis, x = 0
−3y + 12 = 0
−3y = −12
y = 4
B (0,4)
(ii)
–10 –8 –6 –4 –2–2
246
m y
x
–4
00
2 4 6 8 10 12 14
B = (0,4)
A = (–3,0)
k
(iii) k: 4x − 3y + 12 = 0
−3y = −4x − 12
y = 4 __ 3 x + 4
Slope of m = −3 ___ 4
m: y − y1 = m(x − x1)
y − 4 = −3 ___ 4 (x − 0)
4y − 16 = −3x
m: 3x + 4y − 16 = 0
Q. 12. (i) x +2y −10 = 0 – x –+ y
+– 2 = 0
3y − 12 = 0 3y = 12 y = 4 x − y + 2 = 0 x − 4 + 2 = 0 x − 2 = 0 x = 2 R (2,4)
(ii) k: x − y + 2 = 0 Q (3,5): 3 − 5 + 2 = 0? 0 = 0 ∴ Q ∈ k
(iii) |QR| = √_________________
(3 − 2)2 + (5 − 4)2
= √_________
(1)2 + (1)2
= √__
2
Q. 13. (i) x + y − 8 = 0 – x –+ 2y
+– 10 = 0
3y − 18 = 0 3y = 18 y = 6 x + y −8 = 0 x + 6 − 8 = 0 x − 2 = 0 x = 2 R (2,6)
13Active Maths 2 (Strands 1–5): Ch 20 Solutions
(ii) a: x + y − 8 = 0 At x-axis, y = 0
x − 8 = 0
x = 8 Q (8,0)
(iii) b: x −2y + 10 = 0 At y-axis, x = 0
−2y + 10 = 0
−2y = −10
y = 5 P (0,5)
(iv)
–11–10–9 –8 –7 –6 –5 –4 –3 –2 –1
–2–1
10
23456789
0 1 2 3 4 5 6 7 8 9 10 11 12
R = (2,6)
Q = (8,0)
P = (0,5)
b
a
y
x
(v) T ( 0 + 8 ______ 2 , 5 + 0 ______ 2 ) = T ( 4,2 1 _ 2 ) (vi) Slope PT = slope PQ = −5 ___ 8
y-intercept = 5
∴ y = mx + c
y = −5 ___ 8 x + 5
8y = −5x + 40
5x + 8y = 40
Q. 14.
Line Slopep positiveq 0r undefi neds negativet 0
Q. 15. (i) C ( −1 + 5 _______ 2 , −3 −7 _______ 2 ) = C ( 4 __ 2 , −10 ____ 2 ) = C(2,−5)
(ii) Slope AB = −7 + 3 _______ 5 + 1 = −4 ___ 6 = −2 ___ 3
(iii) Slope p = 3 __ 2 c(2,−5)
p: y − y1 = m(x − x1)
y + 5 = 3 __ 2 (x − 2)
2y + 10 = 3x − 6 3x − 2y − 16 = 0
(iv) p: 3x − 2y − 16 = 0 −2y = −3x + 16
y = 3 __ 2 x − 8
(v) y-intercept (0,−8)
Q. 16. (a) Slope = rise ___ run = 3 __ 1 = 3
y-intercept = 2 ∴ y = 3x + 2
(b)
–10 –8 –6 –4 –2
2
4
6
8
10l2y
x
l1
00
2 4 6 8 10 12 14
l3
14 Active Maths 2 (Strands 1–5): Ch 20 Solutions
(c)
–10 –8 –6 –4 –2
2
4
6
8
10l2y
x
l1
00
2 4 6 8 10 12 14
l3
l4(0,4)
(d) l4: y = ( −1 ___ 3 ) x + 4
(27,−4): −4 = ( −1 ___ 3 ) 27 + 4?
−4 = −9 + 4? −4 = −5 × ∴ (27,−4) ∉ l4
Q. 17. (i) Slope of AB = 4 − 0 ______ 0 − 3 = −4 ___ 3
∴ Slope of BC = 3 __ 4
(ii) B(0, 4) C(8, k)
|BC| = √___________________
(x2 − x1)2 + (y2 − y1)
2
= √_________________
(8 − 0)2 + (k − 4)2
= √_________________
64 + k2 − 8k + 16
= √____________
k2 − 8k + 80
(iii) |BC| = 10
√____________
k2 − 8k + 80 = 10
k2 − 8k + 80 = 100
k2 − 8k − 20 = 0
(k + 2)(k − 10) = 0
k = −2 OR k = 10
# as k is positive
∴ k = 10
(iv) |AB| = √_________________
(0 − 3)2 + (4 − 0)2
= √_______
9 + 16 = √
___ 25
= 5 ∴ Area ABCD = 5(10) = 50 units2
Q. 18. (i) Slope AB = 4 − 2 ______ 3 − 1 = 2 __ 2 = 1
(ii) Slope BC = −2 − 4 _______ 9 − 3 = −6 ___ 6 = −1
m1m2 = (1)(−1) = −1 ∴ AB ⊥ BC
(iii) y − y1 = m(x − x1) y − 4 = −1(x −3) y − 4 = −x + 3 x + y − 7 = 0
(iv) |AB| = √_________________
(3 − 1)2 + (4 − 2)2 = √
_______ 22 + 22
= √__
8 = √
__ 4 √
__ 2
= 2 √__
2 ∴ p = 2
(v) |BC| = √__________________
(9 − 3)2 + (−2 −4)2 = √
__________ 62 + (−6)2
= √___
72 = √
_______ 36 × 2
= 6 √__
2 ∴ Area ΔABC = 1 __ 2 ( 2 √
__ 2 ) ( 6 √
__ 2 )
= 6(2) = 12 units2
Q. 19. (a) |BC| = √___________________
(10 − 3)2 + (9 − 10)2
= √_______
49 + 1
= √___
50
= 7.071 km
(b) |BE| = √__________________
(6 − 3)2 + (6 − 10)2
= √_______
9 + 16
= √___
25
= 5
|BE| = |CE| as E is centre of a circle
∴ distance from B to E to C = 10 km
10 km − 7.071 km = 2.929 km
≈ 3 km
(c) Slope AB = 10 − 8.5 ________ 3 − 1 = 1. 5 ____ 2 = 3 __ 4
y − y1 = m(x − x1) y − 10 = 3 __ 4 (x − 3) 4y − 40 = 3x − 9 3x − 4y + 31 = 0 is equation of
Tangent Street
(d) Slope = −4 ___ 3
y − y1 = m(x − x1)
y − 8 = −4 ___ 3 (x − 17)
3y − 24 = −4x + 68
4x + 3y − 92 = 0 is equation of Perpendicular Avenue
15Active Maths 2 (Strands 1–5): Ch 20 Solutions
(e) 3x − 4y + 31 = 0 × 3 4x + 3y − 92 = 0 × 4 9x − 12y + 93 = 0 16x + 12y − 368 = 0 25x − 275 = 0 25x = 275 x = 11 3(11) − 4y + 31 = 0 33 − 4y + 31 = 0 −4y + 64 = 0 −4y = −64 y = 16
(11,16) co-ordinates of museum
(f) From Town Hall (0,0) to intersection of Straight Road and East Drive (1,0)
1 km Then North on Straight Road to
A(1,8.5) 8.5 km Then travel from A on Tangent
Street to Museum (11,16)
√_____________________
(11 − 1)2 + (16 − 8.5)2
√___________
102 + (7.5)2
√____________
100 + 56.25 = √_______
156.25
= 12.5 km
Total distance travelled is 1 + 8.5 + 12.5 = 22 km
Q. 20. (a) A(6,1) B(2, −1)
Slope of AB = −1 − 1 _______ 2 − 6 = −2 ___ −4 = 1 __ 2
y − y1 = m (x − x1)
y + 1 = 1 __ 2 (x − 2)
2y + 2 = x − 2
x − 2y − 4 = 0
(b) At y-axis x = 0
⇒ −2y −4 = 0
−2y = 4
y = −2 C(0,−2)
(c) |AB| = √___________________
(2 − 6)2 + (−1 − 1)2
= √_____________
(−4)2 + (−2)2
= √___
20
= √__
4 √__
5
= 2 √__
5
|AC| = √___________________
(0 − 6)2 + (−2 − 1)2
= √_______
36 + 9
= √___
45
= √__
9 √__
5
= 3 √__
5
∴ |AB|
_____ |AC|
= 2 √__
5 ____ 3 √
__ 5 = 2 __ 3
Q. 21.
Slope Line
2 k
1 __ 8 m
0 l
– 1 __ 4 j
–1 n