Find the area of the triangle …

Let  A → (0, -1), B → (2, 1) and C→ (0, 3) be the vertices of the triangle ABC. Let D, E and F be the mid-points of sides BC, CA and AB respectively. Then, 
 
{tex}D \to \left( {\frac{{2 + 0}}{2},\frac{{1 + 3}}{2}} \right){/tex}
{tex} \Rightarrow D \to \left( {1,2} \right){/tex}
{tex}E \to \left\{ {\frac{{0 + 0}}{2},\frac{{3 + ( - 1)}}{2}} \right\}{/tex}
{tex}\Rightarrow E \to (0,1){/tex}
{tex}F \to \left\{ {\frac{{2 + 0}}{2},\frac{{1 + ( - 1)}}{2}} \right\}{/tex}
{tex}\Rightarrow F \to (1,0){/tex}
{tex}\therefore{/tex} Area of the triangle DEF
{tex}= \frac{1}{2}{/tex}[1(1 - 0) + 0(0 - 2) + 1(2 - 1)]
{tex}= \frac{1}{2}{/tex}[1 + 0 + 1]
= 1 square unit.
Again, area of the triangle ABC
{tex}= \frac{1}{2}{/tex}[0(1 - 3) + 2 {3 - (-1)} + 0(-1 -1)]
= 4 square units
{tex}\therefore{/tex} Ratio of the area of the triangle formed to the area of the given triangle = 1 : 4