Perpendicular Distance from Straight Line in Plane to Point/General Form/Proof 2

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Theorem

Let $\LL$ be a straight line embedded in a cartesian plane, given by the equation:

$a x + b y + c = 0$

Let $P$ be a point in the cartesian plane whose coordinates are given by:

$P = \tuple {x_0, y_0}$


Then the perpendicular distance $d$ from $P$ to $\LL$ is given by:

$d = \dfrac {\size {a x_0 + b y_0 + c} } {\sqrt {a^2 + b^2} }$


Proof

Let a perpendicular be dropped from $P$ to $\LL$ at $Q$.

Let $PQ$ make an angle $\alpha$ with the $x$-axis.

Let $p$ be the length of $PQ$.

Then the coordinates of $Q$ are given by:

$Q = \tuple {x_0 + p \cos \alpha, y_0 + p \sin \alpha}$

$Q$ lies on $a x + b y + c$, and so:

$a \paren {x_0 + p \cos \alpha} + b \paren {y_0 + p \sin \alpha} + c = 0$
\(\ds a \paren {x_0 + p \cos \alpha} + b \paren {y_0 + p \sin \alpha} + c\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds p \paren {a \cos \alpha + b \sin \alpha}\) \(=\) \(\ds -\paren {a x_0 + b y_0 + c}\)

But from Condition for Straight Lines in Plane to be Perpendicular:

\(\ds \tan \alpha \paren {-\dfrac a b}\) \(=\) \(\ds -1\)
\(\ds \leadsto \ \ \) \(\ds \tan \alpha\) \(=\) \(\ds \dfrac b a\)
\(\ds \leadsto \ \ \) \(\ds a \cos \alpha + b \sin \alpha\) \(=\) \(\ds a \dfrac a {\sqrt {a^2 + b^2} } + b \dfrac b {\sqrt {a^2 + b^2} }\)
\(\ds \) \(=\) \(\ds \sqrt {a^2 + b^2}\)
\(\ds \leadsto \ \ \) \(\ds p\) \(=\) \(\ds -\dfrac {a x_0 + b y_0 + c} {\sqrt {a^2 + b^2} }\)

The minus sign has no immediate significance, and the result follows.

$\blacksquare$


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