Question

In a triangle ABC if $A=(1,2)$ and internal angle bisectors through $B$ and $C$ are $y=x$ and $y=-2x$, then the inradius $r$ of the $\Delta ABC$ is

• A. $\frac{1}{\sqrt{3}}$
• B. $\frac{1}{\sqrt{2}}$
• C. $\frac{2}{3}$
• D. None of these

Answer 1

The correct option is B $\frac{1}{\sqrt{2}}$

Let $A_1(h_1,k_1)$ be the image of A about the line $y=x$ which lies on $BC.$
$\frac{h_1-2}{-1}=\frac{k_1-1}{1}=\frac{-2(-1)}{2}$
$\Rightarrow h_1=1,k_1=2$
So, $A_1\equiv (1,2)$
Let $A_2(h_2,k_2)$ be the image of $A$ about the line $y=-2x$ which lies on $BC.$
$\frac{h_2-2}{2}=\frac{k_2-1}{1}=\frac{-2\left(5\right)}{5}$
$\Rightarrow h_2=-2,k_2=1$
So, $A_2\equiv (-2,1)$
Equation of $BC$ is
$y-2=\frac{1}{3}(x-1)$
$x-3y+5=0$
Point of intersection of angle bisectors i.e.$I\equiv (0,0)$

Inradius= $=\left|\frac{C}{\sqrt{a^2+b^2}}\right|$
so,Inradius $=\left|\frac{5}{\sqrt{1+9}}\right|=\frac{5}{\sqrt{10}}=\frac{1}{\sqrt{2}}$
Hence, option 'B' is correct.