Page 47 - ACE YR IGCSE A TOP APPR' TO MATHS

(c) Midpoint 11 (a) Midpoint = ( –10 + 2 , 4 – 5 )
( –1 + 9 3 + 7 ) 2 2
= 2 , 2 1
= (4, 5) (shown) [2] = (–4, – 2 ) [2]
3 – (–5) 8 4 4 – (–5) 9
8 (a) m = = = – (b) m = =
–6 – 0 –6 3 –10 – 2 –12
4 3
y = – x – 5 [3] = –
3 4
(b) Length = √(–6 – 0) + (3 – (–5)) 3
2
2
( )
4 = – (–10) + c
= √36 + 64 4
=10 units [3] c = – 7
(c) R(6, 3) [1] 2
12 × 8 3 7
(d) Area = y = – x – [3]
2 4 2
= 48 square units [2] (c) 3 = – 3 (1) + c
9 (a) Length = √(–8 – 6) + (12 – (–2)) 2 15 4
2
2
2
= √(–14) + (14) c = 4
= 14√2 [3] y = – 3 x + 15 [3]
(b) Midpoint = ( –8 + 6 , 12 + (–2) ) 4 3 4 4
2
2
= (–1, 5) [2] (d) m = –1 ÷ – 4 = 3
(c) Gradient, m = 12 – (–2) = –1 3 = 4 (1) + c
–8 – 6
1
12 = –1(–8) + c 5 3
c = 4 c = 3
y = –x + 4 [2] Equation for the perpendicular line to RS:
(d) m = –1 = 1 4 5
2 –1 y = 3 x + 3
5 = (1)(–1) + c

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7
3
5
4
c = 6 3 x + 3 = – 4 x – 2
y = x + 6 [2] 25x 31
Length × 3 = –
10 (a) = 12 12 6
2
Length of triangle = 8 units x = – 62
p = 6 – 8 units 25
p = –2 [2] y = – 41
2 – 0 1 25
(
(b) (i) m = = 62 41
6 – 2 2 T – 25 , – 25 ) [4]
1
0 = ( ) (2) + c 12 2y = 5 + 3x
2
c = –1 y = 3 x + 5
y = 1 x – 1 [3] 2 2
2 m × m = –1
1
2
1
(ii) q = (–2) – 1 2
2 m = –1 × 3
2
q = –2 [2] 2
1 = –
(c) 5 = (6) + c 3
2 2 2
c = 2 3 = – 3 (2) + c
1
y = x + 2 [3] 2 4
2 c = 3 + 3
1
(d) y = (–2) + 2 = 2
2 2
= 1 y = – 3 x + 2
D(–2, 1) [2] 3y = –2x + 6
3y + 2x – 6 = 0 [6]
Answers 167

Answers.indd 167 15/03/2022 11:08 AM