Question

Find the area of $△ABC$ with vertices $A(0,-1),B(2,1)$ and $C(0,3)$. Also, find the area of the triangle formed by joining the midpoints of its sides. Show that the ratio of the areas of two triangles is $4:1$.

Answer 1

Area$=\frac{1}{2}\left|\begin{array}{c}0\\ -1\end{array}\begin{array}{c}2\\ 1\end{array}\begin{array}{c}0\\ 3\end{array}\begin{array}{c}0\\ -1\end{array}\right|$

$=\frac{1}{2}|2+6|=\frac{8}{2}=4$

Mid-points are D$=\frac{2+0}{2},\frac{3+1}{2}=(1,2)$

$E=\frac{0+0}{2},\frac{3-1}{2}=(0,1)$

$F=\frac{0+2}{2},\frac{1-1}{2}=(1,0)$

Area of $△DEF=\frac{1}{2}\left|\begin{array}{c}1\\ 2\end{array}\begin{array}{c}0\\ 1\end{array}\begin{array}{c}1\\ 0\end{array}\begin{array}{c}1\\ 2\end{array}\right|$

$=\frac{1}{2}|1-1+2|=1$

$\frac{Areaof△ABC}{Areaof△DEF}=\frac{4}{1}=4$