# Straight Line – Point Slope Form

## Point-Slope Form

So far we’ve covered two forms of the equation of a straight line. Here’s another one, known as the point-slope form.

Consider a line which passes through the point (x1, y1), and has the slope m (=tanθ).

I’ll derive it’s equation using two different methods again.

## Method I

We know the slope-intercept form of the line.. lets begin with that. Let the equation be y = mx + c.

Now we haven’t been given the value of c, but we know that the line passes through the point (x1, y1).

Since the relation y = mx + c holds true for all the points on the line, it must also hold true for xand y1.

Therefore, we can write, y= mx+ c. This gives the value of c as y– mx1.

Substituting this back in the equation, and rearranging the terms, we get the equation as y – y1=m(x – x1)

“But I don’t know what the slope intercept form is!”. No problem. We can derive the equation from scratch, as usual. Let’s move on to the next method.

## Method II

Let’s take any point P(x, y) on the line, and do some constructions as below.

Now tanθ=PQ/AQ which implies m=(y – y1)/(x – x1). (I have skipped a lot of explanation here.. go back to see how PQ and AQ are calculated)

And… we get the same equation as before y – y1=m(x – x1)

## Lesson Summary

1. The equation of a line having a given slope m, and passing through a given point (x1, y1), has the equation: y – y1=m(x – x1)

This form will be one of the most frequently used forms, so keep it handy. See you in the next lesson with some examples !