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:

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

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

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


General Form

Let $L_1$ and $L_2$ be straight lines embedded in a cartesian plane, given in general form:

\(\ds L_1: \, \) \(\ds l_1 x + m_1 y + n_1\) \(=\) \(\ds 0\)
\(\ds L_2: \, \) \(\ds l_2 x + m_2 y + n_2\) \(=\) \(\ds 0\)

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

$\tan \psi = \dfrac {l_1 m_2 - l_2 m_1} {l_1 l_2 + m_1 m_2}$


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 by the definition of slope:

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

and so:

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

$\blacksquare$


Also presented as

When presenting Angle between Straight Lines in Plane, some sources retain the form:

$\tan \psi = \dfrac {\tan \psi_2 - \tan \psi_1} {1 + \tan \psi_1 \tan \psi_2}$


Sources

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