Question

A(6, 1), B(8, 2) and C(9, 4) are the vertices of a parallelogram ABCD. If E is the midpoint of DC, find the area of $∆ADE$. [CBSE 2015]

Answer 1

Let (x, y) be the coordinates of D and $\left(x\text{'},y\text{'}\right)$ be the coordinates of E. Since, the diagonals of a parallelogram bisect
each other at the same point, therefore
$$\begin{gathered} \frac{x+8}{2}=\frac{6+9}{2}\Rightarrow x=7 \\ \frac{y+2}{2}=\frac{1+4}{2}\Rightarrow y=3 \end{gathered}$$
Thus, the coordinates of D are (7, 3).
E is the midpoint of DC, therefore
$$\begin{gathered} x\text{'}=\frac{7+9}{2}\Rightarrow x\text{'}=8 \\ y\text{'}=\frac{3+4}{2}\Rightarrow y\text{'}=\frac{7}{2} \end{gathered}$$
Thus, the coordinates of E are $\left(8,\frac{7}{2}\right)$.
Let $A\left(x_1,y_1\right)=A\left(6,1\right),E\left(x_2,y_2\right)=E\left(8,\frac{7}{2}\right)\mathrm{and}D\left(x_3,y_3\right)=D\left(7,3\right)$. Now
$$\begin{gathered} \mathrm{Area}\left(∆ABC\right)=\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_{\mathit{1}}\mathit{-}{y}_{\mathit{2}}\right)\right] \\ =\frac{1}{2}\left[6\left(\frac{7}{2}-3\right)+8\left(3-1\right)+7\left(1\mathit{-}\frac{\mathit{7}}{\mathit{2}}\right)\right] \\ =\frac{1}{2}\left[\frac{3}{2}\right] \\ =\frac{3}{4}\mathrm{sq}.\mathrm{unit} \end{gathered}$$
Hence, the area of the triangle $∆ADE$ is $\frac{3}{4}\mathrm{sq}.\mathrm{units}$.