Bisectors of Angles between Two Straight Lines/Normal Form

Theorem
Let $\LL_1$ and $\LL_2$ be straight lines embedded in a cartesian plane $\CC$, expressed in normal form as:

The angle bisectors of the angles formed at the point of intersection of $\LL_1$ and $\LL_2$ are given by:

Proof
Let $A'SA$ and $B'SB$ be the straight lines $\LL_1$ and $\LL_2$ respectively, intersecting at the point $S$.

Let $P = \tuple {x, y}$ be an arbitrary point on either of the angle bisectors of $\angle ASB$ or $\angle BSA'$.


 * Bisectors-of-angles.png

Drop perpendiculars $PM$ from $P$ to $SA$ and $PN$ from $P$ to $SB$.

Because:
 * $\angle PSM = \angle PSN$
 * $\angle PMS = \angle PNS$
 * $PS$ is common

we have that:
 * $\triangle PSM = \triangle PSN$

and so:
 * $PM = PN$

Having regard only for the magnitude of the perpendiculars:

which is the equation of a straight line in normal form.

This represents either the angle bisector of $\angle ASB$ or the angle bisector of $\angle BSA'$.

Considering now the signs of the perpendiculars, it is seen that while $PM$ and $PM'$ have the same sign, $PN$ and $PN'$ have opposite signs.

Hence:
 * for one bisector we have $PM = PN$
 * for the other bisector we have $PM = -PN$.

So the equation for the other bisector is $x \paren {\cos \alpha + \cos \beta} + y \paren {\sin \alpha + \sin \beta} = p + q$.