Question 7:
--- 7
(i)
(ii) ---
7
(i)
(ii) | A(4, 6), B (10, 2).
M = (7, 4)
m of AB = −2
3
m of perpendicular = 3
2
→ y−4= 3(x−7)
2
Eqn of line parallel to AB through (3, 11)
y−11=−2(x−3)
→
3
Sim eqns → C (9, 7) | B1
B1
M1 A1
[4]
M1
DM1A1
[3] | co
co
Use of m 1 m 2 = −1 & their midpoint
in the equation of a line. co
Needs to use m of AB
Must be using their correct lines.
Co
8 (a)
(b)
(i) | 1st, 2nd, nth are 56, 53 and −22
a = 56, d = −3
−22 = 56 + (n – 1)(−3)
→ n = 27
( ( ))
S 27 = 2 2 7 112+26 −3
→ 459
1 st , 2 nd , 3 rd are 2k + 6, 2k and k + 2.
2k k+2
Either =
2k+6 2k
or uses a, r and eliminates
→2k2 −10k−12=0
→ k = 6 | M1
A1
M1
A1
[4]
M1
DM1
A1
[3] | Uses correct u n formula.
co
Needs positive integer n
Co
Correct method for equation in k.
Forms quad. or cubic equation with
no brackets or fractions.
Co
Page 6 | Mark Scheme | Syllabus | Paper
Cambridge International AS/A Level – May/June 2015 | 9709 | 12
(ii) | S ∞ = 1− a r with r = 2k 2k +6 or k 2 + k 2 (= 2 3 )
→ 54 | M1
A1
[2] | Needs attempt at a and r and S ∞
Co
9
(i)
(ii)
(iii) | OA = 2i + 4j + 4k and OB =3i + j + 4k
OA.OB= 6 + 4 +16 = 26
OA = 36, OB = 26
26
Cos AOB =
6 26
→ 31.8º
 1 
 
AB = b – a = −3
 
0
 
2 3
   
OC =4+2AB or 1+ AB
   
4 4
   
 4 
 
OC =−2
 
4
 
 4 
 
1 −2
Unit vector ÷ modulus → 6 
 
4
 
OC = 6, OA = 6 | M1
M1
M1
A1
[4]
B1
M1
M1 A1
[4]
B1
[1] | Must be numerical at some stage
Product of 2 moduli
All linked correctly
co
Correct link
÷ by modulus. co
co
Page 7 | Mark Scheme | Syllabus | Paper
Cambridge International AS/A Level – May/June 2015 | 9709 | 12