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.

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Solution

Let the coordinates of B and C be $(x_{2}, y_{2})$ and $(x_{3}, y_{3})$ , respectively.

D is the midpoint of AB.

So,

$(1, 0) = (\frac{x_{2} + 0}{2}, \frac{y_{2} - 1}{2}) \Rightarrow 1 = \frac{x_{2}}{2} and 0 = \frac{y_{2} - 1}{2} \Rightarrow x_{2} = 2 and y_{2} = 1$

Thus, the coordinates of B are (2, 1).

Similarly, E is the midpoint of AC.

So,

$(0, 1) = (\frac{x_{3} + 0}{2}, \frac{y_{3} - 1}{2}) \Rightarrow 0 = \frac{x_{3}}{2} and 1 = \frac{y_{3} - 1}{2} \Rightarrow x_{3} = 0 and y_{3} = 3$

Thus, the coordinates of C are (0, 3).

Also, F is the midpoint of BC. So, its coordinates are

$(\frac{2 + 0}{2}, \frac{1 + 3}{2}) = (1, 2)$

Now,

Area of a triangle = $\frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})]$

Thus, the area of $∆$ ABC is

$\frac{1}{2} [0 (1 - 3) + 2 (3 + 1) + 0 (- 1 - 1)] = \frac{1}{2} \times 8 = 4 square units$

And the area of $∆$ DEF is

$\frac{1}{2} [1 (1 - 2) + 0 (2 - 0) + 1 (0 - 1)] = \frac{1}{2} \times (- 2) = 1 square unit (Taking the numerical value, as the area cannot be negative)$

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