Equation of Straight Line in Plane/Point-Slope Form

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Theorem

Let $\LL$ be a straight line embedded in a cartesian plane, given in slope-intercept form as:

$y = m x + c$

where $m$ is the slope of $\LL$.

Let $\LL$ pass through the point $\tuple {x_0, y_0}$.


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

$y - y_0 = m \paren {x - x_0}$


Parametric Form

Let $\LL$ be a straight line embedded in a cartesian plane, given in point-slope form as:

$y - y_0 = \paren {x - x_0} \tan \psi$

where $\psi$ is the angle between $\LL$ and the $x$-axis.


Then $\LL$ can be expressed by the parametric equations:

$\begin {cases} x = x_0 + t \cos \psi \\ y = y_0 + t \sin \psi \end {cases}$


Proof

As $\tuple {x_0, y_0}$ is on $\LL$, it follows that:

\(\ds y_0\) \(=\) \(\ds m x_0 + c\)
\(\ds \leadsto \ \ \) \(\ds c\) \(=\) \(\ds m x_0 - y_0\)

Substituting back into the equation for $\LL$:

\(\ds y\) \(=\) \(\ds m x + \paren {m x_0 - y_0}\)
\(\ds \leadsto \ \ \) \(\ds y - y_0\) \(=\) \(\ds m \paren {x - x_0}\)

$\blacksquare$


Also presented as

This equation can also be seen presented as:

$y - y_0 = \paren {x - x_0} \tan \psi$

where $\psi$ is the angle that $\LL$ makes with the $x$-axis.


Sources

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