A(6, 1), B(8, 2) and C(9, 4) are the vertices of a parallelogram ABCD.

Solution:

Let $(x, y)$ be the coordinates of $D$ and $\left(x^{\prime}, y^{\prime}\right)$ be the coordinates of $E .$ Since, the diagonals of a parallelogram bisect

each other at the same point, therefore

$\frac{x+8}{2}=\frac{6+9}{2} \Rightarrow x=7$

$\frac{y+2}{2}=\frac{1+4}{2} \Rightarrow y=3$

Thus, the coordinates of D are (7, 3).
E is the midpoint of DC, therefore

$x^{\prime}=\frac{7+9}{2} \Rightarrow x^{\prime}=8$

$y^{\prime}=\frac{3+4}{2} \Rightarrow y^{\prime}=\frac{7}{2}$

Thus, the coordinates of $E$ are $\left(8, \frac{7}{2}\right)$.

Let $A\left(x_{1}, y_{1}\right)=A(6,1), E\left(x_{2}, y_{2}\right)=E\left(8, \frac{7}{2}\right)$ and $D\left(x_{3}, y_{3}\right)=D(7,3)$. Now

Area $(\Delta A B C)=\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right]$

$=\frac{1}{2}\left[6\left(\frac{7}{2}-3\right)+8(3-1)+7\left(1-\frac{7}{2}\right)\right]$

$=\frac{1}{2}\left[\frac{3}{2}\right]$

$=\frac{3}{4}$ sq. unit

Hence, the area of the triangle $\Delta A D E$ is $\frac{3}{4}$ sq. units.