Condition for 3 Points in Plane to be Collinear

Theorem
Let $A = \tuple {x_1, y_1}, B = \tuple {x_2, y_2}, C = \tuple {x_3, y_3}$ be points in the Cartesian plane.

Then:
 * $A$, $B$ and $C$ are collinear

the determinant:
 * $\begin {vmatrix}

x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \\ \end {vmatrix}$ equals zero.

Proof
We have that:


 * $A$, $B$ and $C$ are collinear


 * the area of $\triangle ABC = 0$
 * the area of $\triangle ABC = 0$


 * $\dfrac 1 2 \size {\paren {\begin {vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \\ \end {vmatrix} } } = 0$ (from Area of Triangle in Determinant Form)
 * $\dfrac 1 2 \size {\paren {\begin {vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \\ \end {vmatrix} } } = 0$ (from Area of Triangle in Determinant Form)


 * $\begin {vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \\ \end {vmatrix} = 0$
 * $\begin {vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \\ \end {vmatrix} = 0$