Equation of Straight Line through Intersection of Two Straight Lines

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Theorem

Let $\LL_1$ and $\LL_2$ be straight lines embedded in a cartesian plane $\CC$, expressed using the general equations:

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


Let $\LL_3$ be a third straight line embedded in $\CC$, passing through the point of intersection of $\LL_1$ and $\LL_2$.


$\LL_3$ can be expressed using the general equation:

$(1): \quad \paren {l_1 x + m_1 y + n_1} - k \paren {l_2 x + m_2 y + n_2} = 0$


Proof

Let $P = \tuple {x, y}$ be the point of intersection of $\LL_1$ and $\LL_2$.

We have that:

\(\ds l_1 x + m_1 y + n_1\) \(=\) \(\ds 0\)
\(\ds l_2 x + m_2 y + n_2\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds k \paren {l_2 x + m_2 y + n_2}\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds \paren {l_1 x + m_1 y + n_1} - k \paren {l_2 x + m_2 y + n_2}\) \(=\) \(\ds 0\)

and so equation $(1)$:

$\paren {l_1 x + m_1 y + n_1} - k \paren {l_2 x + m_2 y + n_2} = 0$

is satisfied by the point $P$.


Then:

\(\ds \paren {l_1 x + m_1 y + n_1} - k \paren {l_2 x + m_2 y + n_2}\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds \paren {l_1 - k l_2} x + \paren {m_1 - k m_2} y + \paren {n_1 - k n_2}\) \(=\) \(\ds 0\)


Each of $l_1 - k l_2$, $m_1 - k m_2$ and $n_1 - k n_2$ is a real number.

Hence $(1)$ is the equation of a straight line.

$\blacksquare$


Also presented as

This result can also be conveniently presented as follows:

Let $u = l_1 x + m_1 y + n_1$.

Let $v = l_2 x + m_2 y + n_2$.


Let $\LL_1$ be defined by the equation $u = 0$.

Let $\LL_2$ be defined by the equation $v = 0$.


Then the equation of the straight line passing through the point of intersection of $\LL_1$ and $\LL_2$ can be written as:

$u - k v = 0$


Sources

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