Question

Find the area of the triangle formed by joining the midpoints of the sides of the triangle whose vertices are A(2, 2), B(4, 4) and C(2, 6).

Answer 1

Let A(2, 2), B(4, 4) and C(2, 6) be the vertices of a triangle ABC.
Let D,E and F be the midpoints of AB, BC and CA, respectively.
Then, the coordinates of D, E and F​ are:
$D\left(\frac{2+4}{2},\frac{2+4}{2}\right),E\left(\frac{4+2}{2},\frac{4+6}{2}\right)\text{and}F\left(\frac{2+2}{2},\frac{2+6}{2}\right)$
$$\begin{gathered} \text{i.e.}D\left(3,3\right),E\left(3,5\right)\text{and}F\left(2,4\right) \\ \text{Then},\left(x_1=3,y_1=3\right),\left(x_2=3,y_2=5\right)\text{and}\left(x_3=2,y_3=4\right) \end{gathered}$$
$\text{Area of}∆DEF=\frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|$

$$\begin{gathered} =\frac{1}{2}\left|3\left(5-4\right)+3\left(4-3\right)+2\left(3-5\right)\right| \\ =\frac{1}{2}\left|3\left(1\right)+3\left(1\right)+2\left(-2\right)\right| \\ =\frac{1}{2}\left|3+3-4\right| \\ =\left(\frac{1}{2}\times 2\right) \\ =1\text{sq unit} \end{gathered}$$