Question

Find the area of $∆ABC$ with A(1, −4) and midpoints of sides through A being (2, −1) and (0, −1). [CBSE 2015]

Answer 1

Let $\left(x_2,y_2\right)\mathrm{and}\left(x_3,y_3\right)$ be the coordinates of B and C respectively. Since, the coordinates of A are (1, −4), therefore
$$\begin{gathered} \frac{1+x_2}{2}=2\Rightarrow x_2=3 \\ \frac{-4+y_2}{2}=-1\Rightarrow y_2=2 \\ \frac{1+x_3}{2}=0\Rightarrow x_3=-1 \\ \frac{-4+y_3}{2}=-1\Rightarrow y_3=2 \end{gathered}$$
Let $A\left(x_1,y_1\right)=A\left(1,-4\right),B\left(x_2,y_2\right)=B\left(3,2\right)\mathrm{and}C\left(x_3,y_3\right)=C\left(-1,2\right)$. Now
$$\begin{gathered} \mathrm{Area}\left(∆ABC\right)=\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left({\mathit{y}}_{\mathit{1}}\mathit{-}{\mathit{y}}_{\mathit{2}}\right)\right] \\ =\frac{1}{2}\left[1\left(2-2\right)+3\left(2+4\right)-1\left(\mathit{-}4\mathit{-}2\right)\right] \\ =\frac{1}{2}\left[0+18+6\right] \\ =12\mathrm{sq}.\mathrm{units} \end{gathered}$$
Hence, the area of the triangle $∆ABC$ is 12 sq. units.