__5.4 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*+

*c*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:*Given that the equation of a straight line is

*y*= 3 – 4*x*. Find the gradient and*y*-intercept of the line?

*Solution:**y*= 3 – 4

*x*

*y*= – 4

*x*+ 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:*Given that the equation of a straight line is 4

*x*+ 6*y*– 3 = 0. What is the gradient and*y*-intercept of the line?*Solution:*4

*x*+ 6*y*– 3 = 06

*y*= –4*x*+ 3$\begin{array}{l}y=-\frac{2}{3}x+\frac{1}{2}\leftarrow \overline{)y=mx+c}\\ \therefore \text{Gradient}m=-\frac{2}{3}\\ \text{}y-\text{intercept},\text{}c=\frac{1}{2}\end{array}$