Two vertices of a triangle are (2, –1) and (3, 2) and third vertex lies on the line x + y = 5. If the area of the triangle is 4 units then third vertex is

Let A ≡ (2, –1) and B ≡ (3, 2).
Let the third vertex be $C\left(\alpha ,\beta \right)$
Then, $\alpha +\beta =5$ (given)                                          (1)
Area of  $△ABC=\frac{1}{2}\left|\begin{array}{ccc}2& -1& 1\\ 3& 2& 1\\ \alpha & \beta & 1\end{array}\right|=\pm 4$ (given)
$\Rightarrow \beta -3\alpha =1$                                                        (2)
$or \beta -3\alpha =-15$                                                   (3)
Solving (1) and (2), we get, $\alpha$ = 1, $\beta$ = 4
Solving (1) and (3), we get, $\alpha$ = 5, $\beta$ = 0
Thus, the third vertex is either (5, 0) or (1, 4).