The two vertices of triangle are \((2,-1),(3,2)\) and the third vertex lies on \(x+y=5 .\) The area of the triangle is 4 units, then the third vertex is

Since, the third vertex \(\left(x_{1}, y_{1}\right)\) lie on the line \(x+y=5\)

\(\therefore\) \(x_{1}+y_{1} =5\)
\(y_{1} =5-x_{1}\)
\(\therefore\) Coordinate of \(C\) is \(\left(x_{1}, 5-x_{1}\right)\)
Given, area of \(\Delta A B C=4\) units
\(\therefore \frac{1}{2}\left|\begin{array}{ccc}2 & -1 & 1 \\ 3 & 2 & 1 \\ x_1 & 5-x_1 & 1\end{array}\right|=4\)
Using \(R_{2} \rightarrow R_{2}-R_{1} \text { and } R_{3} \rightarrow R_{3}-R_{1},\)
\(\left|\begin{array}{ccc}2 & -1 & 1 \\ 1 & 3 & 0 \\ x_{1} & -2 & 6-x_{1} & 0\end{array}\right|=8\)
\(\Rightarrow 6-x_{1}-3\left(x_{1}-2\right)=\pm 8 \)
\( \Rightarrow 6-x_{1}-3 x_{1}+6=\pm 8 \)
\( \Rightarrow 12-8=4 x_{1} \text { or } 4 x_{1}=20 \)
\( \Rightarrow x_{1}=1 \text { or } x_{1}=5 \)
\( \therefore y_{1}=5-1=4 \text { or } y_{1}=0\)
\(\therefore C\left(x_{1}, y_{1}\right)=C(1,4)\) or \(C(5,0)\)