Page 43 - ACE YR IGCSE A TOP APPR' TO ADD MATH

9. (a) x = 5 – 2y …… (1) Gradient of PQ = Gradient of SR
y + xy = 6 …… (2) 1 = c – 1
2
12 b – –4
Substitute equation (1) into (2): b + 4 = 12c – 12
y + (5 − 2y)(y) = 6
2
y + 5y – 2y – 6 = 0 b = 12c – 16 …… (2)
2
2
y – 5y + 6 = 0 Substitute equation (2) into (1):
2
(y – 3)(y – 2) = 0 9(12c − 16) = 56 + 8c
y = 3 or y = 2 108c − 144 = 56 + 8c
x = –1 or x = 1 100c = 200
P(−1, 3) and Q(1, 2) c = 2
[3] b = 12(2) – 16
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
(b) Length of PQ = 8
= √(–1 – 1) + (3 – 2) [8]
2
2
= √5 unit 8 + 1 9 – 5
[1] 11. (a) Midpoint of AB = 1 2 , 2 2
3 – 2 1
(c) m = = – 9
2
1 –1 – 1 2 = 1 2 , 2
m m = –1
1 2 1 9 – (–5)
m = –1 ÷ – m = 8 – 1
1
2
= 2 2 = 2
,
Midpoint = 1 –1 + 1 3 + 2 2 m m = –1 1
2
1
2
2
2
1
= 0, 5 2 2 m = – 2
y = mx + c y = mx + c 9
1
1
2
5 = 0 + c 2 = – 2 21 2 + c
2 17
c = 5 c =
2 4
1
y = 2x + 5 y = – x + 17
2 [3] 2 4 [5]
1
10. (a) Midpoint = 1 16 + (–4) , 11 + 1 2 (b) 0 = – x + 17
2
2
2
4
= (6, 6) x = 17
[1] 2
2
(b) PS = PQ D 1 17 , 0
2
2
√(–4 – a) + (1 – 10) = √(a – 16) + (10 – 11) 2
2
2
2
2
2
16 + 8a + a + 81 = a – 32a + 256 + 1 CD = 1 17 – (–3) + (0 – (–5))
2
2
40a = 160 2
a = 4 = 12.54 units
[2]
Gradient of PS = 10 – 1 (c) Area
4 – (–4) 8 1 –3 17 8
= 9 = 1 2
8 2 9 –5 –5 0 9
2
Gradient of QP = 11 – 10 (8 × −5) + (1 × −5) + (–3 × 0) + 1 17 × 9 –
16 – 4 = 1 17 2
1
= 1 2 (9 × 1) − (−5 × –3) − −5 × 2 2 − (0 × 8)
12 1 153 85
Gradient of PS = Gradient of QR = 2 –40 – 5 + 2 – 9 – 15 + 2
9 = 11 – c = 1 |50|
8 16 – b 2
144 – 9b = 88 – 8c = 25 unit
2
9b = 56 + 8c …… (1) [3]
Answers 183

Answers Add Math.indd 183 14/03/2022 12:29 PM