Page 140 - Mathematics_F1

y − 3 = − 2

x + 4
Cross multiplication gives,
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y −= − 2(x + 4) Tanzania Institute of Education
3
3
y −= − 2x − 8
y = − 2x − 83+
y = − 2x − 5
Therefore, the equation of the straight line is y = − 2x − 5.

Example 6.10

Find the equation of a straight line which passes through the points A (3, 7)
and B (–2, –3).

Solution
Using the formula for finding an equation of a straight line,

y y− 1 = y − 2 y 1
−
xx x − x
1 2 1
Let ( , ) (3, 7)xy = and ( ,xy ) ( 2, 3).= − − Substitute these values into the
1 1 2 2
equation as follows:
37
y − 7 = −−
23
x − 3 −−
Simplification gives,
y − 7 = − 10
x − 3 − 5

y − 7 = 2
Mathematics Form One
x − 3

Cross multiply and rearrange to get, y = 2x + 1.

Therefore, the equation of the line passing through the given points is y = 2x + 1.

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25/09/2025 15:01:37
Mathematics form 1.indd 133
Mathematics form 1.indd 133 25/09/2025 15:01:37