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.