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

## 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 \ \$ $\ds \begin {vmatrix} x - x_1 & y - y_1 \\ x_2 - x_1 & y_2 - y_1 \end {vmatrix}$ $=$ $\ds 0$ Definition of 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

$\blacksquare$