Question

Find area of a triangle formed by joining mid-points of sides of another triangle whose vertices are (0, -1), (2, 1) and (0, 3). Find ratio of this area to the area of the given triangle.

Answer 1

Vertices of the given triangle are A $(0,-1),B(2,1),C(0,3)$.
Let D, E, F be mid-points of its sides.
Coordinates of D, E, and F are given by
$D=(\frac{0+2}{2},\frac{-1+1}{2})=(1,0)$
$E=(\frac{0+0}{2},\frac{3-1}{2})=(0,1)$
$F=(\frac{2+0}{2},\frac{1+3}{2})=(1,2)$
Area of a triangle
= $\frac{1}{2}\left\{x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right\}$
Area of $\Delta DEF=\frac{1}{2}\left\{1\right(2-1)+1(1-0)+0(0-2\left)\right\}$
$=\frac{1}{2}(1+1)=1squareunits$
Area of $\Delta ABC=\frac{1}{2}\left[0\right(1-3)+2\{3-(-1\left)\right\}+0(-1-1\left)\right]$
$=\frac{1}{2}\left\{8\right\}=4squareunits$
Therefore, the required ratio is 1:4.