Equation of Straight Line in Plane/Slope-Intercept Form

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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.


Sources


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