Bisectors of Adjacent Angles between Straight Lines Meeting at Point are Perpendicular/Proof 2

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Theorem

If a straight line meets another straight line, the bisectors of the two adjacent angles between them are perpendicular.


Proof

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

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


From Bisectors of Angles between Two Straight Lines, the angle bisectors of the angles formed at the point of intersection of $\LL_1$ and $\LL_2$ are given by:

\(\ds x \paren {\cos \alpha - \cos \beta} + y \paren {\sin \alpha - \sin \beta} - \paren {p - q}\) \(=\) \(\ds 0\)
\(\ds x \paren {\cos \alpha + \cos \beta} + y \paren {\sin \alpha + \sin \beta} - \paren {p + q}\) \(=\) \(\ds 0\)

These are in the form $l x + m y + n = 0$.


We use Condition for Straight Lines in Plane to be Perpendicular to prove that $l_1 l_2 + m_1 m_2 = 0$, where:

\(\ds l_1\) \(=\) \(\ds \cos \alpha - \cos \beta\)
\(\ds l_2\) \(=\) \(\ds \cos \alpha + \cos \beta\)
\(\ds m_1\) \(=\) \(\ds \sin \alpha - \sin \beta\)
\(\ds m_2\) \(=\) \(\ds \sin \alpha + \sin \beta\)


Hence

\(\ds l_1 l_2 + m_1 m_2\) \(=\) \(\ds \paren {\cos \alpha - \cos \beta} \paren {\cos \alpha + \cos \beta} + \paren {\sin \alpha - \sin \beta} \paren {\sin \alpha + \sin \beta}\)
\(\ds \) \(=\) \(\ds \paren {\cos^2 \alpha - \cos^2 \beta} + \paren {\sin^2 \alpha - \sin^2 \beta}\) Difference of Two Squares
\(\ds \) \(=\) \(\ds \paren {\cos^2 \alpha + \sin^2 \alpha} - \paren {\cos^2 \beta + \sin^2 \beta}\) rearranging
\(\ds \) \(=\) \(\ds 1 - 1\) Sum of Squares of Sine and Cosine
\(\ds \) \(=\) \(\ds 0\)

Hence the result.

$\blacksquare$


Sources

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