## 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 (x_{1}, y_{1}) and (x_{2}, y_{2})

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 **(x _{1}, y_{1})** and has slope

**m**is given by

**y-y**

_{1}=m(x-x_{1})We could start with this equation, as we atleast know that the line passes through (x_{1}, y_{1}).

But we have no idea about the slope. Instead, we’ve been another point (x_{2}, y_{2}), through which the line passes.

That means, the coordinates (x_{2}, y_{2}) must satisfy the above equation. That is, y_{2}-y_{1}=m(x_{2}-x_{1}), or m=(y_{2}-y_{1})/(x_{2}-x_{1})

And that’s all. The required equation is **y-y _{1}=\(\frac{y_2-y_1}{x_2-x_1}\)(x-x_{1})**

This also tells me one more thing.

The slope of the line joining two points, (x_{1}, y_{1}) and (x_{2}, y_{2}) is given by (y_{2}-y_{1})/(x_{2}-x_{1}). Have a look..

The slope of the line AB is given by tanθ=BC/AC=(y_{2 }– y_{1})/(x_{2 }– x_{1}). 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

- The equation of the line joining two points (x
_{1}, y_{1}) and (x_{2}, y_{2}) is given by y – y_{1}=\(\frac{y_2-y_1}{x_2-x_1}\)(x – x_{1}) - The slope of the line joining two points (x
_{1}, y_{1}) and (x_{2}, y_{2}) is equal to\(\frac{y_2-y_1}{x_2-x_1}\)

See you soon, with some examples.

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