Question

Find area of the triangle whose vertices are $\left(-8,4\right),\left(-6,6\right)$ and $\left(-3,9\right)$

Answer 1

Solve for area of the required triangle

Let the vertices of triangle be,

$$\begin{gathered} A\left(x_1,y_1\right)=\left(-8,4\right) \\ B\left(x_2,y_2\right)=\left(-6,6\right) \\ C\left(x_3,y_3\right)=\left(-3,9\right) \end{gathered}$$

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 $△ABC=\frac{1}{2}\left|-8\left(6-9\right)-6\left(9-4\right)-3\left(4-6\right)\right|$

$=\frac{1}{2}\left|24-30+6\right|=0sq.units$

Hence, area of the triangle with vertices $\left(-8,4\right),\left(-6,6\right)$ and $\left(-3,9\right)$ is $0sq.units$