Area of Quadrilateral in Determinant Form

Theorem
Let $A = \tuple {x_1, y_1}$, $B = \tuple {x_2, y_2}$, $C = \tuple {x_3, y_3}$ and $D = \tuple {x_4, y_4}$ be points in the Cartesian plane.

Let $A$, $B$, $C$ and $D$ form the vertices of a quadrilateral.

The area $\mathcal A$ of $\Box ABCD$ is given by:


 * $\mathcal A = \dfrac 1 2 \paren {\size {\paren {\begin{vmatrix}

x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \\ \end{vmatrix} } } + \size {\paren {\begin{vmatrix} x_1 & y_1 & 1 \\ x_4 & y_4 & 1 \\ x_3 & y_3 & 1 \\ \end{vmatrix} } } }$

Proof
$\Box ABCD$ can be divided into $2$ triangles: $\triangle ABC$ and $\triangle ADC$.

Hence $\mathcal A$ is the sum of the areas of $\triangle ABC$ and $\triangle ADC$.

From Area of Triangle in Determinant Form:

Hence the result.