Perpendicular Distance from Straight Line in Plane to Point/Normal Form
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Theorem
Let $\LL$ be a straight line in the Cartesian plane.
Let $\LL$ be expressed in normal form:
- $x \cos \alpha + y \sin \alpha = p$
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:
- $\pm d = x_0 \cos \alpha = y_0 \sin \alpha - p$
where $\pm$ depends on whether $P$ is on the same side of $\LL$ as the origin $O$.
Proof
First suppose that $P$ is on the opposite side of $\LL$ from the origin $O$.
Let $MP$ be the ordinate of $P$.
Let $N$ be the point of intersection between $\LL$ and the perpendicular through $O$.
Let $ON$ be produced to $N'$ where $PN'$ is the straight line through $P$ parallel to $\LL$.
We have that:
- $d = NN'$
and so:
- $x_0 \cos \alpha + y_0 \sin \alpha = ON' = p + d$
That is:
- $d = x_0 \cos \alpha + y_0 \sin \alpha - p$
By a similar construction, if $P$ is on the same side of $\LL$ as the origin $O$:
- $-d = x_0 \cos \alpha + y_0 \sin \alpha - p$
$\blacksquare$
Sources
- 1933: D.M.Y. Sommerville: Analytical Conics (3rd ed.) ... (previous) ... (next): Chapter $\text {II}$. The Straight Line: $6$. Distance of a point from a line