Condition for Concurrency of Three Straight Lines

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Theorem

Let $3$ straight lines $\LL_1$, $\LL_2$ and $\LL_3$ be embedded in a cartesian plane $\CC$, expressed using the general equations:

\(\ds l_1 x + m_1 y + n_1\) \(=\) \(\ds 0\)
\(\ds l_2 x + m_2 y + n_2\) \(=\) \(\ds 0\)
\(\ds l_3 x + m_3 y + n_3\) \(=\) \(\ds 0\)


Then $\LL_1$, $\LL_2$ and $\LL_3$ are concurrent if and only if:

$\begin {vmatrix} l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \\ l_3 & m_3 & n_3 \end {vmatrix} = 0$

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


Proof

Necessary Condition

Let $\LL_1$, $\LL_2$ and $\LL_3$ be concurrent.

From Equation of Straight Line through Intersection of Two Straight Lines, $\LL_1$, $\LL_2$ and $\LL_3$ are concurrent if and only if $\LL_3$ has an equation of the form:

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

That is:

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


Hence:

\(\ds \begin {vmatrix} l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \\ l_3 & m_3 & n_3 \end {vmatrix}\) \(=\) \(\ds \begin {vmatrix} l_1 & l_2 & l_3 \\ m_1 & m_2 & m_3 \\ n_1 & n_2 & n_3 \end {vmatrix}\) Determinant of Transpose
\(\ds \) \(=\) \(\ds \begin {vmatrix} l_1 & l_2 & \paren {l_1 - k l_2} \\ m_1 & m_2 & \paren {m_1 - k m_2} \\ n_1 & n_2 & \paren {n_1 - k n_2} \end {vmatrix}\) Definition of $\LL_3$
\(\ds \) \(=\) \(\ds \begin {vmatrix} l_1 & l_2 & l_1 \\ m_1 & m_2 & m_1 \\ n_1 & n_2 & n_1 \end {vmatrix}\) Multiple of Column Added to Column of Determinant: adding $k$ times column $2$ to column $3$
\(\ds \) \(=\) \(\ds 0\) Square Matrix with Duplicate Columns has Zero Determinant

$\Box$


Sufficient Condition

Suppose $\begin {vmatrix} l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \\ l_3 & m_3 & n_3 \end {vmatrix} = 0$

Then the rows of this matrix are linearly dependent.


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That is, there exists $a_1, a_2, a_3 \in \R_{\ne 0}$ such that:

\(\ds a_1 \tuple {l_1, m_1, n_1} + a_2 \tuple {l_2, m_2, n_2} + a_3 \tuple {l_3, m_3, n_3}\) \(=\) \(\ds \bf 0\)
\(\ds \leadsto \ \ \) \(\ds \tuple {l_1, m_1, n_1} + \dfrac {a_2} {a_1} \tuple {l_2, m_2, n_2}\) \(=\) \(\ds - \dfrac {a_3} {a_1} \tuple {l_3, m_3, n_3}\)

$\blacksquare$

$\LL_3: l_3 x + m_3 y + n_3 = 0$ is equivalent to $- \dfrac {a_3} {a_1} l_3 x - \dfrac {a_3} {a_1} m_3 y - \dfrac {a_3} {a_1} n_3 = 0$,

which, from above, is in turn equivalent to:

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

with $k = -\dfrac {a_2} {a_1}$.

Hence from Equation of Straight Line through Intersection of Two Straight Lines, $\LL_1$, $\LL_2$ and $\LL_3$ are concurrent.

$\blacksquare$


Sources

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