# 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.