Question

Find area of the triangle whose vertices are $\left(-5,-1\right)$, $\left(3,-5\right)$ and $\left(5,2\right)$

Answer 1

Solve for area of triangle

Let the vertices of triangle be,

$$\begin{gathered} A\left(x_1,y_1\right)=\left(-5,-1\right) \\ B\left(x_2,y_2\right)=\left(3,-5\right) \\ C\left(x_3,y_3\right)=\left(5,2\right) \end{gathered}$$

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

$\therefore$Area of the given triangle$=\frac{1}{2}\left|-5\left(-5-2\right)+3\left(2+1\right)+5\left(-1+5\right)\right|$

$=\frac{1}{2}\left|35+9+20\right|=32sq.units$

Hence, area of the triangle with vertices $(-5,-1),(3,-5)$ and $\mathrm{(}\mathrm{5}\mathrm{,}\mathrm{2}\mathrm{)}$ is $32sq.units$