Equation of Straight Line in Plane/Slope-Intercept Form

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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.


Sources

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