A(6,1), B (8, 2) and C(9, 4) are three vertices

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

Let the fourth vertex of parallelogram be (x, y).

We know that, the diagonals of a parallelogram bisect each other.

$\therefore$ Mid-point of $B D=$ Mid-point of $A C$

$\Rightarrow$ $\left(\frac{8+x}{2}, \frac{2+y}{2}\right)=\left(\frac{6+9}{2}, \frac{1+4}{2}\right)$

$\left[\because\right.$ mid-point of a line segment joining the points $\left(x_{1}, y_{1}\right)$ and $\left.\left(x_{2}, y_{2}\right)=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)\right]$

$\Rightarrow$ $\left(\frac{8+x}{2}, \frac{2+y}{2}\right)=\left(\frac{15}{2}, \frac{5}{2}\right)$

$\therefore$ $\frac{8+x}{2}=\frac{15}{2}$

$\Rightarrow \quad 8+x=15 \Rightarrow x=7$

and $\frac{2+y}{2}=\frac{5}{2}$

$\Rightarrow$ $2+y=5 \Rightarrow y=3$

So, fourth vertex of a parallelogram is $D(7,3)$.

Now, \text { mid-point of side } D C \equiv\left(\frac{7+9}{2}, \frac{3+4}{2}\right)

$E \equiv\left(8, \frac{7}{2}\right)$

$\left[\because\right.$ area of $\triangle A B C$ with vertices $\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)$ and $\left(x_{3}, y_{3}\right)=\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)\right.$ $\left.+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right]$

$\therefore$ Area of $\triangle A D E$ with vertices $A(6,1), D(7,3)$ and $E\left(8, \frac{7}{2}\right)$,

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

$=\frac{1}{2}\left[6 \times\left(\frac{-1}{2}\right)+7\left(\frac{5}{2}\right)+8(-2)\right]$

$=\frac{1}{2}\left(-3+\frac{35}{2}-16\right)$

$=\frac{1}{2}\left(\frac{35}{2}-19\right)=\frac{1}{2}\left(\frac{-3}{2}\right)$

$=\frac{-3}{4}$ 

[but area cannot be negative]

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