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

## Theorem

Let $\LL$ be the straight line in the Cartesian plane such that:

the slope of $\LL$ is $m$
the $y$-intercept of $\LL$ is $c$

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

$y = m x + c$

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

## Proof 1

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

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

We have:

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

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

$\blacksquare$

## Proof 2

By definition, the $y$-intercept of $\LL$ is $\tuple {0, c}$.

We calculate the $x$-intercept $X = \tuple {x_0, 0}$ of $\LL$:

 $\ds m$ $=$ $\ds \dfrac {c - 0} {0 - x_0}$ $\ds \leadsto \ \$ $\ds x_0$ $=$ $\ds -\dfrac c m$

Hence from the two-intercept form of the Equation of straight line in the plane, $\LL$ can be described as:

 $\ds \dfrac x {-c / m} + \dfrac y c$ $=$ $\ds 1$ Definition of Slope of Straight Line $\ds \leadsto \ \$ $\ds y$ $=$ $\ds m x + c$ after algebra

$\blacksquare$

## 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.

## Also known as

The slope-intercept form of the equation of a straight line in the plane is also known as the gradient-intercept form.