Area of Triangle in Determinant Form

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

The area $\mathcal A$ of the triangle whose vertices are at $A$, $B$ and $C$ is given by:


 * $\mathcal A = \dfrac 1 2 \left|{\left({\begin{vmatrix}

x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \\ \end{vmatrix}}\right)}\right|$

Proof

 * AreaOfTriangleComplex.png

Let $A$, $B$ and $C$ be defined as complex numbers in the complex plane.

The vectors from $C$ to $A$ and from $C$ to $B$ are given by:


 * $z_1 = \left({x_1 - x_3}\right) + i \left({y_1 - y_3}\right)$
 * $z_2 = \left({x_2 - x_3}\right) + i \left({y_2 - y_3}\right)$

From Area of Triangle in Terms of Side and Altitude, $\mathcal A$ is half that of a parallelogram contained by $z_1$ and $z_2$.

Thus: