# 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.

Area of the Triangular Region $ABC$ = Area of Trapezoidal Region $ALMC$ + Area of Trapezoidal Region $CMNB$ - 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 comes, 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

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