Question

In $\Delta ABC$, the coordinates of vertex $A$ is $(4,-1)$, and the lines $x-y-1=0$ and $2x-y=3$ are the internal bisectors of the angles $B$ and $C$ respectively. If the radius of the incircle of the triangle $ABC$ is $r$ then the value of [r] is
(where [.] denotes the greater integer function)

Answer 1

We get the incentre from the intersection of the given angle bisectors
Incentre= $(2,1)$
$AI=\sqrt{(4-2{)}^2+(-1-1{)}^2}=2\sqrt{2}$
Now, $AI=rcosec\left(\frac{A}{2}\right)$
(where $r$ is radius of incircle)
Now $\angle A+\angle B+\angle C=\pi$
$\angle \left(\frac{A}{2}\right)+\angle \left(\frac{B}{2}\right)+\angle \left(\frac{C}{2}\right)=\frac{\pi }{2}......\left(1\right)$
In $\Delta BIC$
$\angle BIC+\angle \left(\frac{B}{2}\right)+\angle \left(\frac{C}{2}\right)=\pi$
$\angle BIC=\frac{\pi }{2}+\angle \frac{A}{2}$
(from equation $1$)
$\tan (\frac{\pi }{2}+\frac{A}{2})=\left|\frac{m_2-m_1}{1+m_1{m}_2}\right|$
$\tan (\frac{\pi }{2}+\frac{A}{2})=\left|\frac{1-2}{1+2}\right|$
$\cot \left(\frac{A}{2}\right)=\frac{1}{3}$
So, $r=\frac{AI}{cosec\left(\frac{A}{2}\right)}=\frac{2\sqrt{2}}{\sqrt{1+\frac{1}{9}}}=\frac{6\sqrt{2}}{\sqrt{10}}=\sqrt{\frac{72}{10}}=\sqrt{7.2}$
$\left[r\right]=2$