Equation of Straight Line in Plane/Normal Form/Polar Form
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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 $\alpha$.
Let $\LL$ be defined in normal form:
- $x \cos \alpha + y \sin \alpha = p$
Then $\LL$ can be presented in polar coordinates as:
- $r \map \cos {\theta - \alpha} = p$
Proof
Let $O$ be the origin of the Cartesian plane and the pole of the corresponding polar frame.
Let $OX$ denote the polar axis, coincident with the $x$-axis.
Let $P$ be an arbitrary point on $\LL$, expressed in polar coordinates as $\polar {r, \theta}$.
Let $N$ be the point on $\LL$ where the normal to $\LL$ intersects $\LL$.
We have that $OP$ is at an angle $\theta$ to $OX$ and is of length $r$.
We have that $ON$ is at an angle $\alpha$ to $OX$ and is of length $p$.
Hence $\angle NOP = \theta = \alpha$.
We also have that $\angle ONP$ is a right angle.
Thus:
- $p = r \map \cos {\theta - \alpha}$
$\blacksquare$
Sources
- 1933: D.M.Y. Sommerville: Analytical Conics (3rd ed.) ... (previous) ... (next): Chapter $\text {II}$. The Straight Line: $4$. Special forms of the equation of a straight line: $(4)$ Polar equation