9.1.1 Equation of a Straight Line

9.1.1 Equation of a Straight Line: y = mx + c

1. Given the value of the gradient, m, and the y-intercept, c, an equation of a straight line
y = mx + can be formed.

2. If the equation of a straight line is written in the form y = mx + c, the gradient, m, and the y-intercept, c, can be determined directly from the equation.

Example 1:
Given that the equation of a straight line is y = 3 – 4x. Find the gradient and y-intercept of the line?
 
Solution:
y= 3 – 4x
y= – 4x + 3 ← (y = mx + c)
Therefore, gradient, m = – 4
y-intercept, c = 3


3. If the equation of a straight line is written in the form ax + by + c = 0, change it to the form y = mx + c before finding the gradient and the y-intercept.

Example 2:
Given that the equation of a straight line is 4x + 6y– 3 = 0. What is the gradient and y-intercept of the line?

Solution:
4x + 6y – 3 = 0
6y = –4x + 3

y= 2 3 x+ 1 2 y=mx+c  Gradient m= 2 3  yintercept, c= 1 2
 


9.1.2 Equation of a Straight Line (Sample Questions)
Question 1:
Given that the equation of a straight line is 4x + 6y – 3 = 0. What is the gradient of the line?

Solution:
4x + 6y – 3 = 0
6y  = – 4x + 3
y = 4 x 6 + 3 6 y = 2 3 x + 1 2 y = m x + c g r a d i e n t , m = 2 3  




Question 2:
Given that the equation of a straight line is y = – 7x + 3. Find the y-intercept of the line?

Solution:
y = mx + c, c is y-intercept of the straight line.
Therefore for the straight line y = – 7x + 3,
y-intercept is 3


Question 3
:

 
 
 
 





Find the equation of the straight line MN if its gradient is equal to 3.
 
Solution:
Given m = 3
Substitute m = 3 and (-2, 5) into y = mx + c.
5 = 3 (-2) + c
5 = -6 + c
c = 11
Therefore the equation of the straight line MN is y = 3+ 11 

Leave a Comment