Equation of Straight Line in Plane/Normal Form/Proof 2

Theorem

Let $\LL$ be a straight line such that:

the perpendicular distance from $\LL$ to the origin is $p$

the angle made between that perpendicular and the $x$-axis is $\alpha$.

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

$x \cos \alpha + y \sin \alpha = p$

Let $P = \tuple {x, y}$ be an arbitrary point on $\LL$.

Let $O$ be the origin of the Cartesian plane in which $\LL$ is embedded.

Let $PQ$ be the perpendicular dropped from $P$ to the $x$-axis.

Let $QS$ be the perpendicular dropped from $Q$ to the line $ON$.

Let $PR$ be the perpendicular dropped from $P$ to the line $QS$.

By definition of cosine:

$OS = OQ \cos \alpha$

By definition of sine:

$PR = PQ \sin \alpha$

Then:

$\ds p$

$=$

$\ds ON$

$\ds $

$=$

$\ds OS + SN$

$\ds $

$=$

$\ds OS + PR$

$\ds $

$=$

$\ds OQ \cos \alpha + PQ \sin \alpha$

$\ds $

$=$

$\ds x \cos \alpha + y \sin \alpha$

$\blacksquare$

1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text {III}$. Analytical Geometry: The Straight Line: Equation of a Straight Line: $\tuple {p, \alpha}$ form