Intersection of Straight Lines in Homogeneous Cartesian Coordinate Form

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\LL_1$ and $\LL_2$ be straight lines embedded in a cartesian plane $\CC$.

Let $\LL_1$ and $\LL_2$ be given in homogeneous Cartesian coordinates by the equations:

\(\ds \LL_1: \ \ \) \(\ds l_1 X + m_1 Y + n_1 Z\) \(=\) \(\ds 0\)
\(\ds \LL_2: \ \ \) \(\ds l_2 X + m_2 Y + n_2 Z\) \(=\) \(\ds 0\)


The point of intersection of $\LL_1$ and $\LL_2$ is unique and has homogeneous Cartesian coordinates given by:

$\tuple {X, Y, Z} = \tuple {m_1 n_2 - m_2 n_1, n_1 l_2 - n_2 l_1, l_1 m_2 - l_2 m_1}$

or any multiple of these.


This can also be expressed as:

$\tuple {X, Y, Z} = \tuple {\begin {vmatrix} m_1 & n_1 \\ m_2 & n_2 \end {vmatrix}, \begin {vmatrix} n_1 & l_1 \\ n_2 & l_2 \end {vmatrix} , \begin {vmatrix} l_1 & m_1 \\ l_2 & m_2 \end {vmatrix} }$


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.

So, first let it be the case that $\LL_1$ and $\LL_2$ are not parallel.


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

Let $P = \tuple {X, Y, Z}$ be the point on both $\LL_1$ and $\LL_2$ expressed in homogeneous Cartesian coordinates.

By definition:

\(\ds x\) \(=\) \(\ds \dfrac X Z\)
\(\ds y\) \(=\) \(\ds \dfrac Y Z\)

where $P = \tuple {x, y}$ described in conventional Cartesian coordinates.

From Intersection of Straight Lines in General Form:

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

Hence:

$\dfrac X Z \dfrac 1 {m_1 n_2 - m_2 n_1} = \dfrac Y Z \dfrac 1 {n_1 l_2 - n_2 l_1} = \dfrac 1 {l_1 m_2 - l_2 m_1}$

and so multiplying by $Z$:

$\dfrac X {m_1 n_2 - m_2 n_1} = \dfrac Y {n_1 l_2 - n_2 l_1} = \dfrac Z {l_1 m_2 - l_2 m_1}$


Hence we have:

\(\ds X\) \(=\) \(\ds \dfrac Z {l_1 m_2 - l_2 m_1} \paren {m_1 n_2 - m_2 n_1}\)
\(\ds Y\) \(=\) \(\ds \dfrac Z {l_1 m_2 - l_2 m_1} \paren {n_1 l_2 - n_2 l_1}\)

for arbitrary $Z \in \R$ such that $Z \ne 0$.

So, let $Z = l_1 m_2 - l_2 m_1$.

Thus we have:

\(\ds X\) \(=\) \(\ds m_1 n_2 - m_2 n_1\)
\(\ds Y\) \(=\) \(\ds n_1 l_2 - n_2 l_1\)
\(\ds Z\) \(=\) \(\ds l_1 m_2 - l_2 m_1\)


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

So, let $l_1 m_2 = l_2 m_1$.

Hence:

$Z = 0$

and it is seen that the point of intersection is now the point at infinity.

$\blacksquare$


Sources