Question:
A triangle $A B C$ lying in the first quadrant has two vertices as $\mathrm{A}(1,2)$ and $\mathrm{B}(3,1)$. If $\angle \mathrm{BAC}=90^{\circ}$, and $\operatorname{ar}(\triangle \mathrm{ABC})=5 \sqrt{5}$ sq. units, then the abscissa of the vertex $C$ is:
Correct Option: , 3
Solution:
$\left(\frac{\mathrm{K}-2}{\mathrm{~h}-1}\right)\left(\frac{1-2}{3-1}\right)=-1 \Rightarrow \mathrm{K}=2 \mathrm{~h}$ ..........(1)
$\sqrt{5}|h-1|=10$
$\because[\Delta \mathrm{ABC}]=5 \sqrt{5}$
$\Rightarrow \frac{1}{2}(\sqrt{5}) \sqrt{(h-1)^{2}+(K-2)^{2}}=5 \sqrt{5}$ ........(2)
$\Rightarrow \mathrm{h}=2 \sqrt{5}+1 \quad(\mathrm{~h}>0)$
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