Find the area of $\Delta A B C$ with $A(1,-4)$ and midpoints of sides through $A$ being $(2,-1)$ and $(0,-1)$.
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.