# Straight Line – Two-Point Form

## Two-Point Form

This lesson will cover the two-point form of the equation of a straight line. That is, equation of the line, which passes through two given points, say (x1, y1) and (x2, y2)

This time I’ll illustrate only one method, which will use the point-slope form of the equation. (described here)

We know that equation of the line which passes through the point (x1, y1) and has slope m is given by y-y1=m(x-x1)

We could start with this equation, as we atleast know that the line passes through (x1, y1).

But we have no idea about the slope. Instead, we’ve been another point (x2, y2), through which the line passes.

That means, the coordinates (x2, y2) must satisfy the above equation. That is, y2-y1=m(x2-x1), or m=(y2-y1)/(x2-x1)

And that’s all. The required equation is y-y1=$$\frac{y_2-y_1}{x_2-x_1}$$(x-x1)

This also tells me one more thing.

The slope of the line joining two points, (x1, y1) and (x2, y2) is given by (y2-y1)/(x2-x1). Have a look..

The slope of the line AB is given by tanθ=BC/AC=(y– y1)/(x– x1). This expression will be quite useful as we’ll see later.

The second method (i.e. starting from scratch) is left for you as an exercise.

I hope that, now you’re pretty comfortable with deriving equations to lines which satisfy given conditions.

## Lesson Summary

1. The equation of the line joining two points (x1, y1) and (x2, y2) is given by y – y1=$$\frac{y_2-y_1}{x_2-x_1}$$(x – x1)
2. The slope of the line joining two points (x1, y1) and (x2, y2) is equal to$$\frac{y_2-y_1}{x_2-x_1}$$

See you soon, with some examples.