Angle between Straight Lines in Plane

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Theorem

Let $L_1$ and $L_2$ be straight lines embedded in a cartesian plane, given by the equations:

\(\displaystyle L_1: \ \ \) \(\displaystyle y\) \(=\) \(\displaystyle m_1 x + c_1\)
\(\displaystyle L_2: \ \ \) \(\displaystyle y\) \(=\) \(\displaystyle m_2 x + c_2\)

Then the angle $\psi$ between $L_1$ and $L_2$ is given by:

$\psi = \arctan \dfrac {m_1 - m_2} {1 + m_1 m_2}$


Proof

Angle-between-Straight-Lines.png


Let $\psi_1$ and $\psi_2$ be the angles that $L_1$ and $L_2$ make with the $x$-axis respectively.

Then from Slope of Straight Line is Tangent of Angle with Horizontal:

\(\displaystyle \tan \psi_1\) \(=\) \(\displaystyle m_1\)
\(\displaystyle \tan \psi_2\) \(=\) \(\displaystyle m_2\)

and so:

\(\displaystyle \tan \psi\) \(=\) \(\displaystyle \map \tan {\psi_2 - \psi_1}\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {\tan \psi_2 - \tan \psi_1} {1 + \tan \psi_1 \tan \psi_2}\) Tangent of Difference
\(\displaystyle \) \(=\) \(\displaystyle \dfrac {m_2 - m_1} {1 + m_1 m_2}\) Definition of $m_1$ and $m_2$

$\blacksquare$


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