Exercise 5A

1 a (x1, y1) = (4, 2), (x2, y2) = (6, 3) g (x1, y1) = (−2, −4), (x2, y2) = (10, 2)
y2  y1 3  2 y2  y1 2   4 
 
x2  x1 6  4 x2  x1 10   2 
1 6
 
2 12
1
b (x1, y1) = (−1, 3), (x2, y2) = (5, 4) 
2
y2  y1 43

x2  x1 5   1
h  x1, y1    12 , 2 ,  x2 , y2    43 , 4
1 y2  y1 4  2
 
6 x2  x1 34  12
c (x1, y1) = (−4, 5), (x2, y2) = (1, 2) 2

y2  y1 25 1
4

x2  x1 1   4  8
3

5
i  x1, y1    14 , 12  ,  x2 , y2    12 , 23 
y2  y1 23  12

d (x1, y1) = (2, −3), (x2, y2) = (6, 5) x2  x1 12  14
y2  y1 5   3 1
  6
x2  x1 62 1
4
8 2
 
4 3
2
j (x1, y1) = (−2.4, 9.6), (x2, y2) = (0, 0)
e (x1, y1) = (−3, 4), (x2, y2) = (7, −6) y2  y1 0  9.6
y2  y1 6  4 
 x2  x1 0   2.4 
x2  x1 7   3
9.6
10 
 2.4
10  4
 1
k (x1, y1) = (1.3, −2.2), (x2, y2) = (8.8, −4.7)
f (x1, y1) = (−12, 3), (x2, y2) = (−2, 8) y2  y1 4.7   2.2 
y2  y1 83 
 x2  x1 8.8  1.3
x2  x1 2   12 
1

2

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 y2  y1 2.5 3 (x1, y1) = (5, b), (x2, y2) = (8, 3)
1 k 
x2  x1 7.5 3b
 3
1 85
 3b
3  3
3
l (x1, y1) = (0, 5a), (x2, y2) = (10a, 0) 3  b  9
y2  y1 0  5a b  12

x2  x1 10a  0
5a 4 (x1, y1) = (c, 4), (x2, y2) = (7, 6)
 64 3
10a 
5 7c 4
 2 3
10 
1 7c 4
 3
2 2  7  c
4
m (x1, y1) = (3b, −2b), (x2, y2) = (7b, 2b) 8  37  c
y2  y1 2b   2b  8  21  3c

x2  x1 7b  3b 13  3c
4b 13
 c
4b 3
1 13

3
n (x1, y1) = (p, p2), (x2, y2) = (q, q2)
1
y2  y1 q 2  p 2 4
 3
x2  x1 q p


 q  p  q  p  5 (x1, y1) = (−1, 2d), (x2, y2) = (1, 4)
q p 4  2d 1

q p 1   1 4
4  2d 1

2 (x1, y1) = (3, −5), (x2, y2) = (6, a) 2 4
y2  y1 1
4 2d  
x2  x1 4
a   5  1
So 4 2 d 0
63 4
a5 1
 4 d 2
3 4
 a  5  12
a7

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6 (x1, y1) = (−3, −2), (x2, y2) = (2e, 5) 9 The gradient of AB is:
5   2  43 1
2 
2e   3 42 2
The gradient of AC is:
7
2 73 4
2e  3 
10  2 8
7  2  2e  3  1

7  4e  6 2
4e  1 The gradients are equal and there is a
point in common between the two line
1
e segments so the points can be joined by
4 a straight line.
7 (x1, y1) = (7, 2), (x2, y2) = ( f, 3f ) 10 If the points A(−2a, 5a), B(0, 4a) and
3f 2 C(6a, a) are collinear, then they all lie
4
f 7 on the same straight line.
3 f  2  4( f  7)
The gradient of AB is:
3 f  2  4 f  28
4a  5a a
2  f  28 
0  (2a) 2a
28  2  f
1
f  26 
2
The gradient of AC is:
8 (x1, y1) = (3, −4), (x2, y2) = (−g, 2g) a  5a 4a
2 g   4  
 3 6a  (2a) 8a
g  3 1

2g  4 2
 3
g  3 The gradients are both  12 and there is
2 g  4  3   g  3 a point in common between the two line
segments so the points are collinear.
2 g  4  3g  9
4  g 9
g  5

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