Find the area of the triangle

Given that, the vertices of triangles

Let $\quad\left(x_{1}, y_{1}\right) \rightarrow(-8,4)$

$\left(x_{2}, y_{2}\right) \rightarrow(-6,6)$

and $\quad\left(x_{3}, y_{3}\right) \rightarrow(-3,9)$

We know that, the area of triangle with vertices $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$ and $\left(x_{3}, y_{3}\right)$

$\Delta=\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}+\left(y_{1}-y_{2}\right)\right]$

$\therefore$ $=\frac{1}{2}[-8(6-9)-6(9-4)+(-3)(4-6)]$

$=\frac{1}{2}[-8(-3)-6(5)-3(-2)]=\frac{1}{2}(24-30+6)$

$=\frac{1}{2}(30-30)=\frac{1}{2}(0)=0$

Hence, the required area of triangle is 0.