# Angle between Straight Lines in Plane

## 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 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

Some sources retain the form:

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