Question

A triangle is formed by the line pair $xy+3x-2y-6=0$ and the line $3x-2y-6=0$. The radius of circle inscribed in the triangle is

• A. $(5-\sqrt{13})$
• B. $\frac{5-\sqrt{13}}{2}$
• C. $\frac{6}{\sqrt{13}}$
• D. $\frac{3}{5+\sqrt{13}}$

Answer 1

The correct option is B $\frac{5-\sqrt{13}}{2}$

The given equation of pair of straight lines $xy+3x-2y-6=0$ can be resolve into two lines:
$xy+3x-2y-6=0$
$(x-2)(y+3)=0$
$x=2$ and $y=-3$
It can be observed from the figure that $△ABC$ is the required triangle (right angled)

Inradius of right angled triangle is given by
$r=\frac{\left(AB\right)\left(BC\right)}{AB+BC+AC}$
$=\frac{\left(3\right)\left(2\right)}{3+2+\sqrt{2^2+3^2}}$ ( $AC=\sqrt{2^2+3^2}$)
$=\frac{6(5-\sqrt{13})}{(5+\sqrt{13})(5-\sqrt{13})}$
$=\frac{5-\sqrt{13}}{2}$
Hence, the correct option is 'B'.