Page 140 - Mathematics

Example 6.11

A straight line passes through the points P( 4,5) and Q( 2,1).− − Find its:

(a) Gradient (b) Equation (c) x and y-intercepts

Solution

(a) The gradient of the line is given by:
y − y
m = 2 1
x − x 1
2
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Let ( , ) ( 4, 5)xy = − and ( ,xy 2 ) ( 2, 1).= − It implies that,
1
2
1
15−
m =
− 2 ( 4)− −

= − 4
2
= − 2

Therefore, the gradient of a straight line is 2.−

(b) Let A( , )xy be any point on the line. Take the points P(– 4, 5) and

A( , )xy to determine the equation of the line, that is,
y − 5
m = . But m = − 2.
x − ( 4)−

Thus, 2 =− y − 5
x + 4
− 2 y − 5
=
1 x + 4

Cross multiplication gives,
Mathematics Form One y − y = − 2x − 85+ − 2x − 3.
4)
2(x +
−
5) =
(y −
5 = −
2x −
85+
y =
3
2x −
−
Therefore, the equation of the straight line is y =

134

25/10/2024 09:51:51
Mathematics form 1.indd 134
Mathematics form 1.indd 134 25/10/2024 09:51:51