Point slope form of equation of a straight line is used to find the equation of a straight line, when the slope of the straight line and a point through which the line is passing are given.

Using the above figure, let us derive below the point slope form of equation of a straight line.

In the above figure, consider a straight line L which makes angle θ with the positive direction of the x-axis (i.e. x axis pointing to the right side).

The angle θ is measured by moving anti-clockwise from the x axis to the straight line L which direction is considered as a positive measure of angles made by straight lines with x axis.

Therefore, slope of the straight line L is tanθ, and slope is denoted by the letter m.

So, *m = tanθ*

Again, the straight line is passing through the point Q(x1, y1)

Now, for any point P(x, y) which lies on the straight line L, the *locus* will be the equation of the straight line L.

Now, in triangle PQL,

tanθ = PL/QL = (y – y1)/(x – x1)

but **tanθ** is the slope of the straight line L and since *m = tanθ,*

so, we get

m = (y – y1)/(x – x1)

After cross – multiplying, we get

*y – y1 = m (x – x1)*

this is the equation of a straight line in point slope form.

Example:

Find the equation of a straight line which is parallel to the straight line y = 4x + 5, and which is passing through the point P (2, 3)

**Solution: **

Equation of the straight line y = 4x + 5 is in slope intercept form

**y = mx + c, **so slope** m = 4. **

Again, slopes of two parallel lines are equal, therefore, slope of the required line is also 4.

Substitute **2 in x1, 3 in y1 and 4 in m** in

**point slope form y – y1 = m (x – x1) **to find the equation of the required line, which will be

y – 3 = 4 (x – 2).

On simplifying, equation of the required straight line is

y = 4x – 5

Slope intercept form.