Equation of Straight Line in Plane/Two-Point Form/Determinant Form

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Theorem

Let $\LL$ be a straight line embedded in a cartesian plane, given in two-point form as:

$\dfrac {x - x_1} {x_2 - x_1} = \dfrac {y - y_1} {y_2 - y_1}$


Then $\LL$ can be expressed in the form:

$\begin {vmatrix} x & y & 1 \\ x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \end {vmatrix} = 0$


Proof

\(\ds \frac {x - x_1} {x_2 - x_1}\) \(=\) \(\ds \frac {y - y_1} {y_2 - y_1}\)
\(\ds \leadsto \ \ \) \(\ds \paren {x - x_1} \paren {y_2 - y_1}\) \(=\) \(\ds \paren {x_2 - x_1} \paren {y - y_1}\)
\(\ds \leadsto \ \ \) \(\ds \paren {x - x_1} \paren {y_2 - y_1} - \paren {x_2 - x_1} \paren {y - y_1}\) \(=\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds \begin {vmatrix} x - x_1 & y - y_1 \\ x_2 - x_1 & y_2 - y_1 \end {vmatrix}\) \(=\) \(\ds 0\) Determinant of Order 2
\(\ds \leadsto \ \ \) \(\ds \begin {vmatrix} x & y & 1 \\ x - x_1 & y - y_1 & 0 \\ x_2 - x_1 & y_2 - y_1 & 0 \end {vmatrix}\) \(=\) \(\ds 0\) Determinant with Unit Element in Otherwise Zero Column
\(\ds \leadsto \ \ \) \(\ds \begin {vmatrix} x & y & 1 \\ -x_1 & -y_1 & -1 \\ x_2 - x_1 & y_2 - y_1 & 0 \end {vmatrix}\) \(=\) \(\ds 0\) Multiple of Row Added to Row of Determinant
\(\ds \leadsto \ \ \) \(\ds \begin {vmatrix} x & y & 1 \\ x_1 & y_1 & 1 \\ x_2 - x_1 & y_2 - y_1 & 0 \end {vmatrix}\) \(=\) \(\ds 0\) Determinant with Row Multiplied by Constant
\(\ds \leadsto \ \ \) \(\ds \begin {vmatrix} x & y & 1 \\ x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \end {vmatrix}\) \(=\) \(\ds 0\) Multiple of Row Added to Row of Determinant

$\blacksquare$


Sources

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