Find the area of the triangle formed by the mid points of sides of the triangle whose vertices are $(2, 1)$, $(-2, 3)$, $(4, -3)$.

Let the vertices of the triangle be $\mathrm{A}(2,1), \mathrm{B}(-2,3)$ and $\mathrm{C}(4,-3)$.

Let the mid-points of $\mathrm{AB}, \mathrm{BC}$ and $\mathrm{AC}$ be $\mathrm{P} . \mathrm{Q}$ and R. respectively.

Here,

$\begin{array}{l} \mathrm{P}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right) \\ \mathrm{Q}\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right) \\ \mathrm{R}\left(\mathrm{x}_{3}, \mathrm{y}_{3}\right) \end{array}$

Now,

$\begin{array}{l} P\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=\left(\frac{2-2}{2}, \frac{1+3}{2}\right)=(0,2) \\ Q\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=\left(\frac{-2+4}{2}, \frac{3-3}{2}\right)=(1,0) \\ R\left(\mathrm{x}_{3}, \mathrm{y}_{3}\right)=\left(\frac{4+2}{2}, \frac{-3+1}{2}\right)=(3,-1) \end{array}$

Therefore, area of $\triangle \mathrm{PQR}$,

$\begin{array}{l} =\frac{1}{2}\left[\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right] \\ =\frac{1}{2}[0(0+1)+1(-1-2)+3(2-0)] \\ =\frac{1}{2} \times 3 \\ =\frac{3}{2} \end{array}$

Hence, area of the required triangle is $1.5 \mathrm{sq}$. units.