Equation of Straight Line in Plane/Slope-Intercept Form

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Theorem

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

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


Then $\mathcal L$ can be described by the equation:

$y = m x + c$

where:

\(\displaystyle m\) \(=\) \(\displaystyle -\dfrac {\alpha_1} {\alpha_2}\)
\(\displaystyle c\) \(=\) \(\displaystyle \dfrac {\beta} {\alpha_2}\)

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


Proof

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


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 $\mathcal L$ and is seen to be constant.


The result follows by setting:

\(\displaystyle m\) \(=\) \(\displaystyle -\dfrac {\alpha_1} {\alpha_2}\)
\(\displaystyle c\) \(=\) \(\displaystyle \dfrac {\beta} {\alpha_2}\)

$\blacksquare$


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