Angle between Straight Lines in Plane/General Form

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Theorem

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

From the general equation for the straight line:

\(\ds L_1: \, \) \(\ds y\) \(=\) \(\ds -\dfrac {l_1} {m_1} x + \dfrac {n_1} {m_1}\)
\(\ds L_2: \, \) \(\ds y\) \(=\) \(\ds -\dfrac {l_2} {m_2} x + \dfrac {n_2} {m_2}\)

Hence the slope of $L_1$ and $L_2$ are $-\dfrac {l_1} {m_1}$ and $-\dfrac {l_2} {m_2}$ respectively.

Let $\psi$ be the angle between $L_1$ and $L_2$, as suggested.

Then:

\(\ds \tan \psi\) \(=\) \(\ds \dfrac {\paren {-\dfrac {l_2} {m_2} } - \paren {-\dfrac {l_1} {m_1} } } {1 + \paren {-\dfrac {l_1} {m_1} } \paren {-\dfrac {l_2} {m_2} } }\) Angle between Straight Lines in Plane
\(\ds \) \(=\) \(\ds \dfrac {\dfrac {l_1} {m_1} - \dfrac {l_2} {m_2} } {1 + \dfrac {l_1} {m_1} \dfrac {l_2} {m_2} }\) simplifying to eliminate unnecessary minus signs
\(\ds \) \(=\) \(\ds \dfrac {l_1 m_2 - l_2 m_1} {l_1 l_2 + m_1 m_2}\) multiplying top and bottom by $m_1 m_2$

$\blacksquare$


Sources