Let $C\left(4, \alpha \right)$ be on $x=4$.

Now, midpoint of AC is $\left(\frac{5}{2}, \frac{\alpha +2}{2}\right)$ which lies on $x+y=5$.

$\therefore \frac{\alpha +2}{2}=\frac{5}{2}\Rightarrow \alpha =3$

$\therefore C\left(4, 3\right)$

Similarly, let $B\left(\beta \right)\left(, 5-\beta \right)$ be on $x+y=5$.

The midpoint of AB $=\left(\frac{\beta +1}{2}, \frac{7-\beta }{2}\right)$ which lies on $x=4\Rightarrow \beta =7\Rightarrow B\left(7, -2\right)$.

$\therefore$ Area of $\mathrm{\Delta }=\frac{1}{2}\left|\begin{array}{ccc}1& 2& 1\\ 7& -2& 1\\ 4& 3& 1\end{array}\right|$

$=\frac{1}{2}\left[1\left(-2-3\right)-2\left(7-4\right)+1\left(21+8\right)\right]=\frac{1}{2}\left[18\right]=9$ sq. units