Page 57 - ACE YR IGCSE A TOP APPR' TO MATHS

→ → 11 11
(c) RT = k QO = – p + r [2]
1 5 5
1( )
– 1 p = k – 2 p → → →
3 3 (c) RS = PS – PR
k = 1 11 11
1 2 = – 5 p + 5 r – (–p + r)
→ → 6 6
RT = k QO = – 5 p + 5 r
2
2( )
– 1 s = k – 1 s PR:RS
6 3 –p: – 6 p
5
k = 1 PR:RS = 5:6 [3]
2
2
k = 1 → → →
2 (d) TR = TO + OR
)
(
Scalar multiples of one another. = – 1 11 r – 6 p + r
→ → 3 5 5
Vectors RT and QO are parallel. [2] = 4 r + 6 p
→ → → 15 15
11 (a) (i) OS = OQ + QS → →
= q + 3t [1] OQ = mTR
6
→ 2 3 p = m ( )
p
(ii) OP = q [1] 5 15
5 m = 1.5
→ → →
→
→
(b) PR = PO + OR OQ = nTR
→ 2 →
4
= PO + OS 2 r = n ( )
r
5 5 15
= – 2 q + 2 (q + 3t) n = 1.5
5 5 Scalar multiples of one another.
6
= t → →
5 Vectors TR and OR are parallel. [4]
6
Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
k =
[2]
5 Transformations
(c) PR :QS
–11
6 1 ( ) [2]
t :3t 11
5
6 :15 2 p(2, 3) [2]
Area of ΔOPR:Area of ΔOQS 3 (a), (b) and (c)
36:225 [2] y
(d) Area of ΔOPR:Area of ΔOQS 8
20 cm :125 cm 2 7
2
Area of PRSQ 6
= 125 – 20 5
= 105 cm 2 [2] 4 Q
3
→ → → 2
12 (a) (i) PR = PO + OR S 1 P
= –p + r [1] x
→ → → –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8
(ii) OQ = OP + PQ R –1
–2
→ 2 → –3
= OP + PR
5 –4
2 –5
= p + (– p + r)
5 –6
–7
= 3 p + 2 r [2] –8
5 5 [7]
→ → →
(b) PS = PO + OS
= – p + 11 r – 6 p
5 5

Answers 177

Answers.indd 177 15/03/2022 11:08 AM