Question

Find area of the triangle formed by joining the midpoints of the sides of the triangle
whose vertices are A(2, 1), B(4, 3) and C(2, 5).

Answer 1

The vertices of the triangle are A(2, 1), B(4, 3) and C(2, 5).
$$\begin{gathered} \mathrm{Coordinates}\mathrm{of}\mathrm{midpoint}\mathrm{of}AB=P\left(x_1,y_1\right)=\left(\frac{2+4}{2},\frac{1+3}{2}\right)=\left(3,2\right) \\ \mathrm{Coordinates}\mathrm{of}\mathrm{midpoint}\mathrm{of}BC=Q\left(x_2,y_2\right)=\left(\frac{4+2}{2},\frac{3+5}{2}\right)=\left(3,4\right) \\ \mathrm{Coordinates}\mathrm{of}\mathrm{midpoint}\mathrm{of}AC=R\left(x_3,y_3\right)=\left(\frac{2+2}{2},\frac{1+5}{2}\right)=\left(2,3\right) \end{gathered}$$
Now
$$\begin{gathered} \mathrm{Area}\mathrm{of}∆PQR=\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right] \\ =\frac{1}{2}\left[3\left(4-3\right)+3\left(3-2\right)+2\left(2-4\right)\right] \\ =\frac{1}{2}\left[3+3-4\right]=1\mathrm{sq}.\mathrm{unit} \end{gathered}$$
Hence, the area of the required triangle is 1 sq. unit.