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