Question

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

Answer 1

Solution: Area of $\Delta ABC=\frac{1}{2}\left|\begin{array}{ccc}3-(-5)& 2-6& \\ -5-8& 6-3& \end{array}\right|$
$=\frac{1}{2}\left|\begin{array}{ccc}8& -4& \\ -13& 3& \end{array}\right|$
$=\, \displaystyle \frac{1}{2}$$\left|\begin{array}{c}8\left(3\right)-4\left(13\right)\end{array}\right|$$\, =\, \displaystyle \frac{1}{2}\begin{vmatrix}24-52\end{vmatrix}\, =\, 14 sq. units.$
Hence the area of the midpoints of the sides $\Delta ABC$ = 1/4 (Area of $\Delta ABC$) = 1/4 (14) = 3.5 sq. units