**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*+

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

**9.1.2 Equation of a Straight Line (Sample Questions)**

**Question 1**:

Given that the equation of a straight line is 4

*x*+ 6*y*– 3 = 0. What is the gradient of the line?

*Solution:*4

*x*+ 6*y*– 3 = 06

$\begin{array}{l}y=-\frac{4x}{6}+\frac{3}{6}\\ y=-\frac{2}{3}x+\frac{1}{2}\\ y=mx+c\\ gradient,\text{}m=-\frac{2}{3}\end{array}$
*y*= – 4*x*+ 3**Question 2**:

Given that the equation of a straight line is

*y*= – 7*x*+ 3. Find the*y*-intercept of the line?

Solution:Solution:

*y*= m

*x*+ c, c is y-intercept of the straight line.

Therefore for the straight line

*y*= – 7*x*+ 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*= 3Substitute

*m*= 3 and (-2, 5) into*y*= m*x*+ c.5 = 3 (-2) + c

5 = -6 + c

c = 11

Therefore the equation of the straight line

*MN*is*y*= 3*x*+ 11