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 $\AA$ of $\Box ABCD$ is given by:

$\AA = \dfrac 1 2 \paren {\size {\paren {\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \\ \end{vmatrix} } } + \size {\paren {\begin{vmatrix} x_1 & y_1 & 1 \\ x_4 & y_4 & 1 \\ x_3 & y_3 & 1 \\ \end{vmatrix} } } }$

Proof

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

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

 $\ds \map \Area {\triangle ABC}$ $=$ $\ds \dfrac 1 2 \size {\paren {\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \\ \end{vmatrix} } }$ $\ds \map \Area {\triangle ADC}$ $=$ $\ds \dfrac 1 2 \size {\paren {\begin{vmatrix} x_1 & y_1 & 1 \\ x_4 & y_4 & 1 \\ x_3 & y_3 & 1 \\ \end{vmatrix} } }$

Hence the result.

$\blacksquare$

Examples

Vertices at $\tuple {2, -1}, \tuple {4, 3}, \tuple {-1, 2}, \tuple {-3, -2}$

Let $Q$ be a quadrilateral embedded in the cartesian plane with vertices at $\tuple {2, -1}$, $\tuple {4, 3}$, $\tuple {-1, 2}$ and $\tuple {-3, -2}$.

The area of $Q$ is given by:

$\map \Area Q = 18$