Question

Find the area of the triangle formed by the midpoints of the sides of $△ABC$ where $A=(3,2),B=(-5,6)$ and $C=(8,3)$

Answer 1

Area of $△ABC$ $=\frac{1}{2}\left|\begin{array}{cc}3-(-5)& 2-6\\ -5-8& 6-3\end{array}\right|$
=$\frac{1}{2}\left|\begin{array}{cc}8& -4\\ -13& 3\end{array}\right|$
=$\frac{1}{2}|8\left(3\right)-4\left(13\right)|$= $\frac{1}{2}$ $|24-52|=14$sq.units
Hence the area of triangle formed by the midpoints of the sides of $△ABC=\frac{1}{4}$ (Area of $△ABC$)=$\frac{1}{4}\left(14\right)=3.5$sq.units