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-triangle

The area of the triangular region $$ABC$$ = the area of trapezoidal region $$ALMC$$ + the area of trapezoidal region $$CMNB$$  the area of trapezoidal region $$ALNB$$
\[\begin{gathered} \Delta ABC = \frac{1}{2}\left( {LA + MC} \right)\left( {LM} \right) + \frac{1}{2}\left( {MC + NB} \right)\left( {MN} \right) – \frac{1}{2}\left( {LA + NB} \right)\left( {LN} \right) \\ \Rightarrow \Delta ABC = \frac{1}{2}\left( {{y_1} + {y_3}} \right)\left( {{x_3} – {x_1}} \right) + \frac{1}{2}\left( {{y_3} + {y_2}} \right)\left( {{x_2} – {x_3}} \right) – \frac{1}{2}\left( {{y_1} + {y_2}} \right)\left( {{x_2} – {x_1}} \right) \\ \Rightarrow \Delta ABC = \frac{1}{2}\left( {{x_3}{y_1} – {x_1}{y_3} + {x_2}{y_3} – {x_3}{y_2} – {x_2}{y_1} + {x_1}{y_2}} \right) \\ \Rightarrow \Delta ABC = \frac{1}{2}\left( {{x_1}{y_2} – {x_1}{y_3} – {x_2}{y_1} + {x_2}{y_3} + {x_3}{y_1} – {x_3}{y_2}} \right) \\ \end{gathered} \]
\[\Delta ABC = \frac{1}{2}\left[ {{x_1}\left( {{y_2} – {y_3}} \right) – {x_2}\left( {{y_1} – {y_3}} \right) + {x_3}\left( {{y_1} – {y_2}} \right)} \right]\]

This formula can be written in the determinant form as follows:
\[\Delta ABC = \frac{1}{2}\left| {\begin{array}{*{20}{c}} 1&1&1 \\ {{x_1}}&{{x_2}}&{{x_3}} \\ {{y_1}}&{{y_2}}&{{y_3}} \end{array}} \right|\]

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
\[\left| {\begin{array}{*{20}{c}} 1&1&1 \\ {{x_1}}&{{x_2}}&{{x_3}} \\ {{y_1}}&{{y_2}}&{{y_3}} \end{array}} \right| = 0\]