Page 147 - Mathematics

6. A line has an equation ax + (a − 1)y + 2a = 0.
(a) Rewrite the equation in the form of y = mx + . c
(b) If the line has y-intercept at (0, 2), find the value of a.
(c) Using the value of a obtained in (b), find the gradient of the line.

Equation of a straight line when given a gradient and a point Tanzania Institute of Education
The equation of a straight line can be determined when given a gradient m and a
)
point A( ,xy lying on the line. Choosing another point P( , )xy which lie on the
1
1
y − y
line, it follows that, m = 1
x x− 1
Cross multiplication gives,
y y− 1 = mx x− ( 1 )
This is the equation of a line with gradient m and passes through the point A( ,xy 1 )
1

Example 6.17

3
A straight line has a gradient of and passes through the point (–1, 2). Find:
4
(a) Its equation in the form y = mx c+
(b) The x and y-intercepts.

Solution
(a) The equation of this line is y y− 1 = mx x− ( 1 ).

3
Given m = and the point ( , )xy = ( 1, 2).−
1
1
4
Substitute the values in the equation to get,
3
y – 2 = ( x – (–1))
4

4(y − 2) 3(x= + 1)

4y = 3x + 11
3 11
y = x + Mathematics Form One
4 4
3 11
Therefore, the equation of the straight line is y = x + .
4 4

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25/10/2024 09:51:55
Mathematics form 1.indd 141 25/10/2024 09:51:55
Mathematics form 1.indd 141