In \[∆\] ABC , the coordinates of vertex A are (0, - 1) and D (1,0) and E(0,10) respectively the mid-points of the sides AB and AC . If F is the mid-points of the side BC , find the area of \[∆\] DEF.
Solution
Let the coordinates of B and C be \[\left( x_2 , y_2 \right)\] and \[\left( x_3 , y_3 \right)\] , respectively.
D is the midpoint of AB.
So,
\[\left( 1, 0 \right) = \left( \frac{x_2 + 0}{2}, \frac{y_2 - 1}{2} \right)\]
\[ \Rightarrow 1 = \frac{x_2}{2} \text { and } 0 = \frac{y_2 - 1}{2}\]
\[ \Rightarrow x_2 = 2 \text{ and } y_2 = 1\]
Thus, the coordinates of B are (2, 1).
Similarly, E is the midpoint of AC.
So,
\[\left( 0, 1 \right) = \left( \frac{x_3 + 0}{2}, \frac{y_3 - 1}{2} \right)\]
\[ \Rightarrow 0 = \frac{x_3}{2} \text{ and } 1 = \frac{y_3 - 1}{2}\]
\[ \Rightarrow x_3 = 0 \text{ and } y_3 = 3\]
Thus, the coordinates of C are (0, 3).
Also, F is the midpoint of BC. So, its coordinates are \[\left( \frac{2 + 0}{2}, \frac{1 + 3}{2} \right) = \left( 1, 2 \right)\]
\[\frac{1}{2}\left[ 0\left( 1 - 3 \right) + 2\left( 3 + 1 \right) + 0\left( - 1 - 1 \right) \right]\]
\[ = \frac{1}{2} \times 8\]
\[ = 4 \text{ square units } \]
And the area of \[∆\] DEF is
\[\frac{1}{2}\left[ 1\left( 1 - 2 \right) + 0\left( 2 - 0 \right) + 1\left( 0 - 1 \right) \right]\]
\[ = \frac{1}{2} \times \left( - 2 \right)\]
\[ = 1 \text{ square unit } \left( \text{ Taking the numerical value, as the area cannot be negative } \right)\]
