# Equation of Straight Line in Plane/Slope-Intercept Form

## Theorem

Let $\LL$ be the straight line defined by the general equation:

$\alpha_1 x + \alpha_2 y = \beta$

Then $\LL$ can be described by the equation:

$y = m x + c$

where:

 $\ds m$ $=$ $\ds -\dfrac {\alpha_1} {\alpha_2}$ $\ds c$ $=$ $\ds \dfrac {\beta} {\alpha_2}$

such that $m$ is the slope of $\LL$ and $c$ is the $y$-intercept.

## Proof

 $\ds \alpha_1 x + \alpha_2 y$ $=$ $\ds \beta$ $\ds \leadsto \ \$ $\ds \alpha_2 y$ $=$ $\ds y_1 - \alpha_1 x + \beta$ $\text {(1)}: \quad$ $\ds \leadsto \ \$ $\ds y$ $=$ $\ds -\dfrac {\alpha_1} {\alpha_2} x + \dfrac {\beta} {\alpha_2}$

Setting $x = 0$ we obtain:

$y = \dfrac {\beta} {\alpha_2}$

which is the $y$-intercept.

Differentiating $(1)$ with respect to $x$ gives:

$y' = -\dfrac {\alpha_1} {\alpha_2}$

By definition, this is the slope of $\LL$ and is seen to be constant.

The result follows by setting:

 $\ds m$ $=$ $\ds -\dfrac {\alpha_1} {\alpha_2}$ $\ds c$ $=$ $\ds \dfrac {\beta} {\alpha_2}$

$\blacksquare$

## Also presented as

This equation can also be seen presented as:

$y = x \tan \psi + c$

where $\psi$ is the angle that $\LL$ makes with the $x$-axis.