Question

In ∆ABC, the coordinates of vertex A are (0, –1) and D (1, 0) and E (0, 1)  respectively the mid-points of the sides AB and AC. If F is the mid-point of side BC, find the area of ∆DEF.

Answer 1

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  \[\left( \frac{2 + 0}{2}, \frac{1 + 3}{2} \right) = \left( 1, 2 \right)\]

Now,

Area of a triangle =  \[\frac{1}{2}\left[ x_1 \left( y_2 - y_3 \right) + x_2 \left( y_3 - y_1 \right) + x_3 \left( y_1 - y_2 \right) \right]\]

Thus, the area of \[∆\]ABC is

\[\frac{1}{2}\left[ 0\left( 1 - 3 \right) + 2\left( 3 + 1 \right) + 0\left( - 1 - 1 \right) \right]\]

\[ = \frac{1}{2} \times 8\]

\[ = 4 \text{ square units } \]

And the area of  \[∆\]DEF is

\[\frac{1}{2}\left[ 1\left( 1 - 2 \right) + 0\left( 2 - 0 \right) + 1\left( 0 - 1 \right) \right]\]

\[ = \frac{1}{2} \times \left( - 2 \right)\]

\[ = 1 \text{ square unit }  \left( \text{Taking the numerical value, as the area cannot be negative} \right)\]