Page 56 - ACE YR IGCSE A TOP APPR' TO MATHS

→ –4 → → →
(b) QR = ( ) (c) OM = OA + AM
1
→
→ → = a + 1 AC
PS = QR 2
→ → → 1 → →
OS = OP + PS = a + 2 (AO + OC)
9
)
–4
= ( ) ( ) = a + 1 ( –a + b + 2 b
+
8
1
3
2
( ) 1 5
5
= 9 [2] = 2 a + 6 b [2]
→ ( ) → → →
2
6 (a) OQ = 1 [1] (d) MB = MO + OB
)
→ 4 = – ( 1 a + 5 b + b
(b) TR = ( ) [1] 1 2 1 6
0
→ → = – 2 a + 6 b [2]
(c) UR = 2 × UO → → →
→ → 9 (a) (i) UR = UP + PR
= 2 × (UT + TO) = a + 4b [1]
[( ) ( )]
2
0
→ → →
= 2 × –2 + 1 (ii) PS = PU + US
( ) 3 →
2
= 2 × –1 = –a + 2 PR
( ) 3
4
= –2 [3] = –a + 2 (4b)
→ → → = –a + 6b [2]
7 (a) AC = AD + DC → → →
= –x + y [1] (iii) RS = RU + US
→ → → = –a – 4b + 6b
(b) AB = AE + EB = –a + 2b [2]
1 → → → → →
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EB + EB
y =

(b) TS = TR + RS
3
4 → 1 → →
EB = y = UR + RS
3 4
→ 3y = 1 (a + 4b) – a + 2b
EB = 4
4 3
→ → → = – a + 3b [2]
CE = CB + BE 4
3y → → →
= x – [3] 10 (a) (i) SP = SO + OP
4 = –s + p [1]
→ → → → → →
(c) DE = DA + AE (ii) TP = TO + OP
1 → 1
= x + EB = − s + p [2]
3 2
3y
→ → →
= x + 1 ( ) (b) (i) RT = RS + ST
3 4
→ →
= x + y [2] = 1 PS + ST
4 3
→ → → 1 ( )
1
8 (a) BA = BO + OA = 3 (s – p) + – 2 s
= –b + a [1] 1 1
→ → → = – 3 p – 6 s [2]
(b) AC = AO + OC → → →
→ → → (ii) QO = QP + PO
= AO + OB + BC 1 → →
= –a + b + 2 b = 3 SP + PO
5 3 1
= –a + b [2] = (–s + p) + (–p)
3 3
= – 1 s – 2 p [2]
3 3
Cambridge IGCSE
TM
176 Ace Your Mathematics

Answers.indd 176 15/03/2022 11:08 AM