Cambridge International AS and A Level Mathematics Pure Mathematics 1

(Michael S) #1
Chapter

(^2)
287
P1^
(ii) AB:BC: CD:
AD:
5
12
5
12
1
3
4
3
,–






,,

(iii) AB = 13; BC = 13; CD = 40 ;
AD = 10


(iv) AB: 5x − 12 y = 0;
BC: 5x + 12 y − 120 = 0;
CD: x − 3 y + 30 = 0;
AD: 4x + 3 y = 0


(v) 90 square units


●?^ (Page^ 58)^
Attempting to solve the equations
simultaneously gives 3 = 4 which is
clearly false so there is no point of
intersection. The lines are parallel.


Exercise 2D (Page 58)


  1 (i) A(1, 1); B(5, 3); C(−1, 10)


(ii) BC = AC = 85


  2 (i)


(ii) (−3, 3)


(iii) 2 x − y = 3; x − 2 y = 0


(iv) (−6, −3); (5, 7)


  3 (i) y=^12 x + 1, y = − 2 x^ + 6


(ii) Gradients =^12 and − 2 ⇒ AC
and BD are perpendicular.
Intersection = (2, 2) = mid-
point of both AC and BD.


(iii) AC = BD = 20


(iv) Square


  4 (i)

(ii) A: (4, 0), B: (0, 11), C: (2, 10)
(iii) 11 square units
(iv) (−2, 21)
  5 (i) (2, 4)
(ii) (0, 3)
  6 (i) –, 21 34 ,–,^1243 parallelogram
(ii) 10
(iii) –^43 , 4x + 3 y = 20
(iv) (4.4, 0.8)
(v) 40 square units
  7 (i) − 3
(ii) x − 3 y + 5 = 0
(iii) x = 1
(iv) (1, 2)
(vi) 3.75 square units
  8 (i) 21 (− 2 + 14) = 6

(ii) gradient of AD = (^8) h
gradient of CD = (^128) – h
(iii) x co-ordinate of D = 16
x co-ordinate of B = − 4
(iv) 160 square units
  9 M(4, 6), A(−8, 0), C(16, 12)
10 (i) 3 x + 2 y = 31
(ii) (7, 5)
11 (i) 2 x + 3 y = 20
(ii) C(10, 0), D(14, 6)
12 (6.2, 9.6)
13 (i) (4, 6)
(ii) (6, 10)
(iii) 40.9 units
14 B(6, 5), C(12, 8)
●?^ (Page^ 63)^
Even values of n: all values of y are
positive; y axis is a line of symmetry.
Odd values of n: origin is the centre
of rotational symmetry of order 2.
Exercise 2E (Page 68)
 1
 2
3
 4
 5
O
9
2 x – y = –9
x – 2y = –9
x
y
–9
4 –^12
–4–^12
O
2
x+2y=22
x+y=2
x
y
4 $ 22
% &
–4
y
3 x
–1 2
y
x
20
(^124)
–3
y
1 5 x
–15
y
3 x
±1 
y
x


Free download pdf