Bisectors of Angles between Two Straight Lines/Normal Form

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Theorem

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

\(\ds \LL_1: \ \ \) \(\ds x \cos \alpha + y \sin \alpha\) \(=\) \(\ds p\)
\(\ds \LL_2: \ \ \) \(\ds x \cos \beta + y \sin \beta\) \(=\) \(\ds q\)


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

\(\ds x \paren {\cos \alpha - \cos \beta} + y \paren {\sin \alpha - \sin \beta}\) \(=\) \(\ds p - q\)
\(\ds x \paren {\cos \alpha + \cos \beta} + y \paren {\sin \alpha + \sin \beta}\) \(=\) \(\ds p + q\)


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:

\(\ds PM\) \(=\) \(\ds x \cos \alpha + y \sin \alpha - p\)
\(\ds PN\) \(=\) \(\ds x \cos \beta + y \sin \beta - q\)
\(\ds \leadsto \ \ \) \(\ds x \paren {\cos \alpha - \cos \beta} + y \paren {\sin \alpha - \sin \beta}\) \(=\) \(\ds p - q\)

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 $x \paren {\cos \alpha + \cos \beta} + y \paren {\sin \alpha + \sin \beta} = p + q$.

$\blacksquare$


Sources