# Intersection of Straight Lines in General Form

## Theorem

Let $\LL_1$ and $\LL_2$ be straight lines embedded in a cartesian plane $\CC$, given by the equations:

 $\ds \LL_1: \ \$ $\ds l_1 x + m_1 y + n_1$ $=$ $\ds 0$ $\ds \LL_2: \ \$ $\ds l_2 x + m_2 y + n_2$ $=$ $\ds 0$

The point of intersection of $\LL_1$ and $\LL_2$ has coordinates given by:

$\dfrac x {m_1 n_2 - m_2 n_1} = \dfrac y {n_1 l_2 - n_2 l_1} = \dfrac 1 {l_1 m_2 - l_2 m_1}$

This point exists and is unique if and only if $l_1 m_2 \ne l_2 m_1$.

### Determinant Form

This result can also be expressed in the form:

$\dfrac x {\begin {vmatrix} m_1 & n_1 \\ m_2 & n_2 \end {vmatrix} } = \dfrac y {\begin {vmatrix} n_1 & l_1 \\ n_2 & l_2 \end {vmatrix} } = \dfrac 1 {\begin {vmatrix} l_1 & m_1 \\ l_2 & m_2 \end {vmatrix} }$

where $\begin {vmatrix} \cdot \end {vmatrix}$ denotes a determinant.

## Proof

First note that by the parallel postulate $\LL_1$ and $\LL_2$ have a unique point of intersection if and only if they are not parallel.

From Condition for Straight Lines in Plane to be Parallel, $\LL_1$ and $\LL_2$ are parallel if and only if $l_1 m_2 = l_2 m_1$.

$\Box$

Let the equations for $\LL_1$ and $\LL_2$ be given.

Let $\tuple {x, y}$ be the point on both $\LL_1$ and $\LL_2$.

We have:

 $\ds l_1 x + m_1 y + n_1$ $=$ $\ds 0$ $\ds \leadsto \ \$ $\ds y$ $=$ $\ds -\dfrac {l_1} {m_1} x - \dfrac {n_1} {m_1}$ $\ds l_2 x + m_2 y + n_2$ $=$ $\ds 0$ $\ds \leadsto \ \$ $\ds y$ $=$ $\ds -\dfrac {l_2} {m_2} x - \dfrac {n_2} {m_2}$ $\ds \leadsto \ \$ $\ds \dfrac {l_1} {m_1} x + \dfrac {n_1} {m_1}$ $=$ $\ds \dfrac {l_2} {m_2} x + \dfrac {n_2} {m_2}$ substituting for $y$ $\ds \leadsto \ \$ $\ds l_1 m_2 x + n_1 m_2$ $=$ $\ds l_2 m_1 x + n_2 m_1$ multiplying by $m_1 m_2$ $\ds \leadsto \ \$ $\ds x \paren {l_1 m_2 - l_2 m_1}$ $=$ $\ds n_2 m_1 - n_1 m_2$ rearranging $\ds \leadsto \ \$ $\ds \dfrac x {m_1 n_2 - m_2 n_1}$ $=$ $\ds \dfrac 1 {l_1 m_2 - l_2 m_1}$ dividing by $\paren {l_1 m_2 - l_2 m_1} \paren {m_1 n_2 - m_2 n_1}$

Similarly:

 $\ds l_1 x + m_1 y + n_1$ $=$ $\ds 0$ $\ds \leadsto \ \$ $\ds x$ $=$ $\ds -\dfrac {m_1} {l_1} y - \dfrac {n_1} {l_1}$ $\ds l_2 x + m_2 y + n_2$ $=$ $\ds 0$ $\ds \leadsto \ \$ $\ds x$ $=$ $\ds -\dfrac {m_2} {l_2} y - \dfrac {n_2} {l_2}$ $\ds \leadsto \ \$ $\ds \dfrac {m_1} {l_1} y + \dfrac {n_1} {l_1}$ $=$ $\ds \dfrac {m_2} {l_2} y + \dfrac {n_2} {l_2}$ substituting for $x$ $\ds \leadsto \ \$ $\ds m_1 l_2 y + n_1 l_2$ $=$ $\ds m_2 l_1 y + n_2 l_1$ multiplying by $m_1 m_2$ $\ds \leadsto \ \$ $\ds y \paren {m_2 l_1 - m_1 l_2}$ $=$ $\ds n_1 l_2 - n_2 l_1$ rearranging $\ds \leadsto \ \$ $\ds \dfrac y {n_1 l_2 - n_2 l_1}$ $=$ $\ds \dfrac 1 {l_1 m_2 - l_2 m_1}$ dividing by $\paren {l_1 m_2 - l_2 m_1} \paren {n_1 l_2 - n_2 l_1}$

$\blacksquare$