Question

Find the area of the triangle formed by joining the mid points of the sides of the triangle
whose vertices are $(0,-1),(2,1)$ and $(0,3)$ .

Answer 1

Let $(x_1,y_1),(x_2,y_2)$ and $(x_3,y_3)$ be the mid-points of $△ABC$

$\Rightarrow x_1=\frac{0+2}{2}=1,y_1=\frac{-1+1}{2}=0$

$\Rightarrow x_2=\frac{2+0}{2}=1,y_2=\frac{1+3}{2}=2$

$\Rightarrow x_3=\frac{0+0}{2}=0,y_3=\frac{3-1}{2}=1$

$\therefore$Mid-points are $(1,0),(1,2)$ and $(0,1)$

Area of the triangle joining the midpoints$=\frac{1}{2}\left|\begin{array}{ccc}1& 0& 1\\ 1& 2& 1\\ 0& 1& 1\end{array}\right|$

$=\frac{1}{2}[1(2-1)-0+1(1-0)]=\frac{1}{2}[1-0+1]=\frac{2}{2}=1sq.unit$