Area of Triangle with Three Vertices

Let $A\left( {{x_1},{y_1}} \right)$, $B\left( {{x_2},{y_2}} \right)$ and $C\left( {{x_3},{y_3}} \right)$ be the vertices of the triangular region as shown in the given diagram. Draw perpendiculars from the points $A$, $B$ and $C$ on the X-axis at the points $L$, $M$ and $N$ respectively. There are three trapezoidal regions formed in this way.

The area of the triangular region $ABC$ = the area of trapezoidal region $ALMC$ + the area of trapezoidal region $CMNB$ - the area of trapezoidal region $ALNB$

This formula can be written in the determinant form as follows:

This gives the area of the triangular region. The negative sign should be omitted if it occurs, as the area should be positive.

NOTE: If the three points $A$, $B$ and $C$ are collinear points (lying on the same line), then no triangular region will form and the area will be zero, so the condition for three points $A\left( {{x_1},{y_1}} \right)$, $B\left( {{x_2},{y_2}} \right)$ and $C\left( {{x_3},{y_3}} \right)$ to be collinear is that