Page 151 - Mathematics
P. 151
0 = −−
2( 2) c+
c = − 4
From y = mx + , c it implies that,
y = − 2x − 4
Therefore, the equation of the straight line is y = − 2x − 4. Tanzania Institute of Education
Example 6.23
Find the gradient and y-intercept in each of the following lines:
(a) 12x + 4y − 15 = 0 (b) 24 8x− − 3y = 0
Solution
(a) Given 12x + 4y − 15 = 0.
Make y the subject of the formula. Thus,
4y = − 12x + 15
Dividing by 4 both sides of the given equation gives,
15
y = − 3x +
4
3
Compare this equation with y mx c= + gives m = − and c = 15 . Thus,
4
15
0, is the y-intercept.
4
Therefore, the gradient is –3 and the y-intercept is 0, 15 .
4
(b) Given 24 8x− − 3y = 0.
Make y the subject of the formula:
− 3y = 8x − 24.
Dividing by –3 on both sides gives,
8
y = − x + 8
3
8
Compare this equation with y = mx c+ to get m = − and c = 8. Mathematics Form One
8 3
Therefore, the gradient is − and the y -intercept is (0, 8).
3
145
25/10/2024 09:51:58
Mathematics form 1.indd 145 25/10/2024 09:51:58
Mathematics form 1.indd 145