Indian Institute of Foreign Trade 2010 Solved Paper

Let us assume that X co-ordinate of the 3rd vertex is a. Then $\mathit{y}=\mathit{a}+5$. So the co-ordinates of the vertex $=\left(\mathit{a},\mathit{a}+5\right)$
Therefore,$6=\frac{1}{2}\times |\begin{array}{ccc}1& 1& 1\\ 4& -1& 1\\ \mathit{a}& \mathit{a}+5& 1\end{array}|$
$\Rightarrow \pm 12=5\mathit{a}+10$
$\Rightarrow \mathit{a}=\frac{2}{5},\frac{-22}{5}$
Hence, the third vertex can be out of the two values $\left(\frac{2}{5},\frac{27}{5}\right)$ or $\left(\frac{22}{5},\frac{3}{5}\right)$
Therefore, option A is the correct answer.