Area of Triangle from Vertices

Area of triangle with vertices (x₁,y₁), (x₂,y₂), (x₃,y₃).

Let the vertices be $A(x_1,y_1)$, $B(x_2,y_2)$, $C(x_3,y_3)$.

Translate so $A$ is at the origin: $B' = (x_2-x_1, y_2-y_1)$ and $C' = (x_3-x_1, y_3-y_1)$.

The area of the triangle $OB'C'$ is half the magnitude of the cross product of $\overrightarrow{OB'}$ and $\overrightarrow{OC'}$:

$$\Delta = \frac{1}{2}\left|(x_2-x_1)(y_3-y_1) - (x_3-x_1)(y_2-y_1)\right|$$

Expanding:

$$\Delta = \frac{1}{2}\left|x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)\right|$$

This is exactly the expansion of:

$$\boxed{\Delta = \frac{1}{2}\left|\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}\right|}$$

The signed version (without absolute value) is positive when the vertices are ordered anticlockwise and negative when clockwise. This sign convention is useful in problems involving orientation.

> > $$ > \Delta = \frac{1}{2}\,|\overrightarrow{AB} \times \overrightarrow{AC}| > $$ > > where $\times$ is the 2D cross product (scalar z-component). This is the direct definition of area via the parallelogram law — the triangle is half the parallelogram spanned by $\overrightarrow{AB}$ and $\overrightarrow{AC}$. > > In 3D, for a triangle with vertices in space, the same formula holds with the full 3D cross product giving a vector whose magnitude is twice the area: > > $$ > \Delta = \frac{1}{2}\,|\overrightarrow{AB} \times \overrightarrow{AC}| > $$ > > The determinant form is simply the coordinate expansion of this cross product.