Question

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

Answer 1

Let $A_1(h_1,k_1)$ be the image of A about the line $y=x$ which lies on BC.

$\frac{h_1-x_1}{a}=\frac{k_1-y_1}{b}=\frac{-2(a{x}_1+b{y}_1+c)}{a^2+b^2}$

$\frac{h_1-1}{-1}=\frac{k_1-2}{1}=-\frac{2\left(1\right)}{2}$

$\Rightarrow h_1=2,k_1=1$

$\text{So},A_1\equiv (2,1)$

Let $A_2(h_2,k_2)$be the image of A about the line $y=-2x$ which lies on BC.

$\frac{h_2-1}{2}=\frac{k_2-2}{1}=-\frac{2\left(4\right)}{5}$

$\Rightarrow h_2=-\frac{11}{5},k_2=\frac{2}{5}$

$So,A_2\equiv (-\frac{11}{5},\frac{2}{5})$

Equation of BC is
$y-1=m(x-2)$

$m=\frac{\frac{2}{5}-1}{-\frac{11}{5}-2}=\frac{2-5}{-11-10}=\frac{1}{7}$

$\Rightarrow y-1=\frac{1}{7}(x-2)$

$\Rightarrow 7y-7=x-2$

$\Rightarrow x-7y+5=0$

Point of interesection of angle bisectors $i.e.I\equiv (0,0)$

$\text{Inradius}=\left|\frac{a{x}_1+b{y}_1+c{z}_1+d}{\sqrt{a^2+b^2}}\right|$

$=\left|\frac{5}{\sqrt{1+49}}\right|$

$=\frac{1}{\sqrt{2}}$
hence, option 'B' correct.