What is the Point Slope Form of a Line Equation

Here you will learn what is the point slope form of a line equation with proof and examples.

Let’s begin – 

What is the Point Slope Form of a Line ?

The equation of a line which passes through the point $P(x_1, y_1)$ and has the slope ‘m’ is

$y – y_1$ = m$(x – x_1)$

Proof :

Let $Q(x_1, y_1)$ be the point through which the line passes and let  P(x, y) be any point on the line.

Then the slope of the line is $y – y_1\over x – x_1$

but, m is the slope of line 

$\therefore$ m = $y – y_1\over x – x_1$ $\implies$ $y – y_1$ = m($x – x_1$)

Hence, $y – y_1$ = m$x – x_1$ is the required equation of the line.

Example : Find the equation of a line passing through (2, -3) and inclined at an angle of 135 with the positive direction of x-axis.

Solution : Here, m = slope of the line = tan 135 = tan(90 + 45) = -cot 45 = -1

$x_1$ = 2, $y_1$ = -3

So, the equation of the line is

$y – y_1$ = m($x – x_1$)

i.e. y – (-3) = -1(x – 2)

$\implies$ y + 3 = -x + 2 $\implies$ x + y + 1 = 0.

which is the required equation of line.

Example : Determine the equation of line through the point (-4, -3) and parallel to x-axis

Solution : Here, m = slope of the line = 0,

$x_1$ = -4, $y_1$ = -3

So, the equation of the line is

$y – y_1$ = m($x – x_1$)

i.e. y + 3 = 0(x + 4)

$\implies$ y + 3 = 0.

which is the required equation of line.