Equation of Straight Line in Plane/Slope-Intercept Form/Proof 1
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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
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:
\(\ds m\) | \(=\) | \(\ds -\dfrac {\alpha_1} {\alpha_2}\) | ||||||||||||
\(\ds c\) | \(=\) | \(\ds \dfrac {\beta} {\alpha_2}\) |
$\blacksquare$
Sources
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- 1933: D.M.Y. Sommerville: Analytical Conics (3rd ed.) ... (previous) ... (next): Chapter $\text {II}$. The Straight Line: $4$. Special forms of the equation of a straight line: $(1)$ Gradient forms