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
- 1933: D.M.Y. Sommerville: Analytical Conics (3rd ed.) ... (previous) ... (next): Chapter $\text {II}$. The Straight Line: $5$