A triangle ABC lying in the first quadrant has two vertices at $A(1,2)$ and $B(3,1)$. If $\mathrm{\angle }BAC={90}^{\circ }$and area $\mathrm{\Delta }\left(ABC\right)=5\sqrt{5}$ sq. units then the abscissa of the vertex C is

Let the coordinates of vertex C be $(\alpha ,\beta )$. It is given that $\mathrm{\angle }BAC={90}^{\circ }$

$\Rightarrow$ Slope of AB x Slope of AC = -1

$\Rightarrow \frac{1-2}{3-1}\times \frac{\beta -2}{\alpha -1}=-1$

$\Rightarrow 2\alpha =\beta$              …(i)

Area of $\mathrm{\Delta }\left(ABC\right)=5\sqrt{5}$

$\begin{array}{l}\Rightarrow \frac{1}{2}AB\times AC=5\sqrt{5}\\ \Rightarrow \sqrt{(3-1{)}^2+(1-2{)}^2}\times \sqrt{(\alpha -1{)}^2+(\beta -2{)}^2}=10\sqrt{5}\\ \Rightarrow \sqrt{5}\sqrt{(\alpha -1{)}^2+(2\alpha -2{)}^2}=10\sqrt{5}\end{array}$

$\begin{array}{l}\Rightarrow \sqrt{5}|\alpha -1|=10\\ \Rightarrow |\alpha -1|=2\sqrt{5}\Rightarrow \alpha -1=\pm 2\sqrt{5}\Rightarrow \alpha =1\pm 2\sqrt{5}\end{array}$