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.
$\blacksquare$
Sources
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics: $\text {III}$. Analytical Geometry: The Straight Line: The length of the perpendicular