Page 42 - ACE YR IGCSE A TOP APPR' TO ADD MATH

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5. Area = x x x x 1 AC = √(x – 3) + (y – 0)
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2 y y y y 1 1 = x – 6x + 9 + y
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x – 6x + y + 8 = 0 …… (1)
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5.5 = 1 –2 d 1
2 3 –1 4 3 2x – 2 = y
11 = |−1 − 8 + 3d − (−6)−(−d) − 4)| 3 2
11 = |–7 + 4d| 4x – 12 = 3y
4x – 12
= 1 −7 + 4d = 11 y = 3 …… (2)
2 d = 4.5 Substitute equation (2) into (1):
= √3 –7 + 4d = –11 4x – 12 2
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d = –1 x – 6x + 3 3 4 + 8 = 0
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= √(7) – [2] 9x – 54x + 16x – 96x + 144 + 72 = 0
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6. y = x – 9x – 10 25x – 150x + 216 = 0
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= 1 8 + 3 dy = 2x – 9 x = –b ± √b – 4ac
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1 dx 2a
= 2 2 Gradient of the tangent line at point A: –(–150) ± √(–150) – 4(25)(216)
2
= m m = −1 = 2(25)
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1
m = –1 ÷ 1 18 12
1 2 x = or x =
= –2 5 5
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2x – 9 = –2 y = or y = – 4
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x =
2 C 1 18 4 2 (invalid, as it is not on AB) or
,
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5
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y = 1 2 2 – 9 1 2 – 10 C 1 12 –4 2 .
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2
,
= –117 5 5 [10]
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Equation of the normal line: (b) m m = –1
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2
y = mx + c m = –1 ÷ 4
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–117 = 1 7 + c 3 3
1 2
4 2 2 = – 4
c = –31
4
1 2
1 – = – 3 12 + c
y = x – 31 5 4 5
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[5] c = 1
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7. (a) Point A: y = – x + 1
4
2x – 2 = 0 [2]
3 2
2x = 2 8. (a) m = –10 – 3
1
3 0 – 7
x = 3 = 1
A(3, 0) m m = –1
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2
m = –1
2
Point B: y = mx + c
2(0) – 2 = y –3 = (–1)(7) + c
3 a c = 4
y = –2a \ y = –x + 4
B(0, –2a) [2]
1 0 0 7 0
Distance AB: (b) Area =
5 = √(3 – 0) + (0 – (–2a)) 2 0 –10 –3 0
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2
1
5 = √9 + 4a = |–70|
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2
25 – 9 = 4a = 35 unit
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2
a = 2 or a = –2 (invalid) [2]
Cambridge IGCSE
TM
182 Ace Your Additional Mathematics
Answers Add Math.indd 182 14/03/2022 12:29 PM