Page 59 - ACE YR IGCSE A TOP APPR' TO ADD MATH

→ → → 2 1 5 10 2
2 1
2
(b) OD = OA + AD = a + 1 – a + d + µ – a + d – d
→ 19 → 3 3 3 9 3
= OA + AB 1 1 10 1
1
2
16 = a + d + µ – a – d
→ 19 → → 9 3 9 3
= OA + (AO + OB)
16 1 1 10 1
1
2
→ 19 → 19 → l(–a + d) = a + d + µ – a – d
= OA – OA + OB 9 3 9 3
16 16 1 10
3 → 19 → –l = – 9 µ
9
= – OA + OB
1
16 16 l = 10 µ –
= – 3 (–3i + 12j) + 19 (13i – 20j) 9 9
1
1
16 16 l = – µ
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= 16i – 26j 3 3
1
1
1
[4] 10 µ – = – µ
→ → → → 9 9 3 3
9. AB = AD + DC + CB 13 µ = 4
= 4a + 3b – a – 3a – b 9 9
= 2b µ = 4
→ → → 13
BD = BC + CD 3
= 3a + b + a – 3b l = 13
= 4a – 2b [9]
→ → → → → →
AE = l(AB + BC) 11. (a) AB = AO + OB
= l(2b + 3a + b) = –a + b
= 3la + 3lb [1]
→ → → 1 →
BE = µBD (b) (i) OE = OD
2
→
→
= µ(BC + CD) 1 → →
= µ(3a + b + a – 3b) = 2 (OA + AD)
= 4µa – 2µb 1 → →
→ → → = 2 (OA + kAB)
AB = AE + EB 1
2b = 3la + 3lb + 2µb – 4µa = 2 [a + k(–a + b)]
3l – 4µ = 0 1
3l + 2µ = 2 = 2 [a – ka + kb]
k
Solve the simultaneous equations = 1 – k a + b
–6µ = –2 2 2 [3]
µ = 1 → → →
3 (ii) OE = OC + CE
→
→
1
1
3l + 2 1 2 = 2 = jOA + CB
3
3
→
→
→
1
l = 4 = jOA + (CO + OB)
9
[8] 3
→
→
→
1
→ → → = jOA + (–jOA + OB)
10. AE = l(AO + OD) 3
= l(–a + d) = 2 ja + b
1
→ → → → 3 3
AE = AB + BC + CE [3]
2 → 1 → → k 1
= OA + BD + µCO (c) =
3 3 2 3
2 → 1 → → → → 2
= OA + (BO + OD) + µ(CD + DO) k =
3 3 3
2 → 1 5 → → 2 → → 2
2 1
2
= OA + 1 – OA + OD + µ BD – OD 1 –
2
3 3 3 3 3 = j
2 → 1 5 → → 2 3
2
= OA + 1 – OA + OD + 1
3 3 3 j = 4
5 →
→
→
2
4
2
µ 3 1 – OA + OD – OD [2]
3
3
Answers 199
Answers Add Math.indd 199 14/03/2022 12:29 PM