Solve this

Let $\left(x_{2}, y_{2}\right)$ and $\left(x_{3}, y_{3}\right)$ be the coordinates of $B$ and $C$ respectively. Since, the coordinates of $A$ are $(1,-4)$, therefore

$\frac{1+x_{2}}{2}=2 \Rightarrow x_{2}=3$

$\frac{-4+y_{2}}{2}=-1 \Rightarrow y_{2}=2$

$\frac{1+x_{3}}{2}=0 \Rightarrow x_{3}=-1$

$\frac{-4+y_{3}}{2}=-1 \Rightarrow y_{3}=2$

Let $A\left(x_{1}, y_{1}\right)=A(1,-4), B\left(x_{2}, y_{2}\right)=B(3,2)$ and $C\left(x_{3}, y_{3}\right)=C(-1,2)$. Now

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

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

$=\frac{1}{2}[0+18+6]$

$=12$ sq. units

Hence, the area of the triangle $\Delta A B C$ is 12 sq. units.