Question

Let three lines $L_1,L_2$ and $L_3$ belonging to the family $x-2y+6+\lambda (x-y+2)=0$ where $\lambda$ is a parameter, be interior angle bisectors of $△ABC.$ If the equation $x+3y-4=0$ represents side $AB$ of the triangle, then the value of $\left[\frac{△}{r\cot \frac{A}{2}+a+r\cot \frac{B}{2}+b+r\cot \frac{C}{2}+c}\right]$ is
(Note: Symbols used have usual meanings in $△ABC$ and $[.]$ denotes the greatest integer function.)

Answer 1

As $\tan \frac{A}{2}=\frac{r}{s-a}$
$\Rightarrow s=r\cot \frac{A}{2}+a$
Similarly, $s=r\cot \frac{B}{2}+b$ and $s=r\cot \frac{C}{2}+c$
$\frac{△}{\sum (r\cot \frac{A}{2}+a)}=\frac{△}{3s}=\frac{r}{3}$

Incentre is the intersection point of $x-2y+6=0$ and $x-y+2=0$ i.e., $(2,4)$
$r=$ distance of $I(2,4)$ from $AB:x+3y-4=0$
Using distance formula of point from line
$r=\left|\frac{2+12-4}{\sqrt{10}}\right|=\sqrt{10}$
$\therefore \left[\frac{r}{3}\right]=1$