Page 149 - Mathematics

1 2
1
y −= − x +
5 5
Thus,
y = − 1 x + 7
5 5

x + 5y − 70= Tanzania Institute of Education
Therefore, the equation of the straight line is x + 5y − 7 0.=

Example 6.19

Find the equation of a line in the form ax by c+ += 0, passing through the
3
point (3, 1) with gradient − .
5
Solution
Given ( , ) (3, 1),xy = m = − 3 .
1
1
5
From y y− 1 = mx x− ( 1 ), it follows that,
3
1
y −= − (x − 3)
5
5y −= − 3x + 9
5
Therefore, the equation of the line is 3x + 5y − 14 0.=

Example 6.20

Find the equation of a straight line in the formax by c+ += 0passing through
the point(3, − 4)having the same gradient as the straight line 5x − 2y = 3.

Solution
Given ( , ) (3,xy = − 4) and 5x − 2y = 3. Rearranging the given equation in
1
1
5 3
the form y = mx c+ gives, y = x − .
2 2
Comparing with y = mx + , c it implies that m = 5
2
From y y− 1 = mx x− ( 1 ), it follows that, Mathematics Form One
5
y − ( 4)− = (x − 3)
2

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25/10/2024 09:51:56
Mathematics form 1.indd 143
Mathematics form 1.indd 143 25/10/2024 09:51:56