Question

If ${}^{\prime }{G}^{\prime }$ is centroid and ${}^{\prime }{I}^{\prime }$ is incentre of the triangle formed by line $3x+4y=12$ with co-ordinate axis and $GI=\frac{2}{3}\sqrt{{\lambda }^2-4}(\lambda >0)$ then $\left[\lambda \right]=$
(where $[.]$ is G.I.F)

Answer 1

For the above triangle,
Centroid $G=(\frac{4}{3},1),$ Incentre$I=(1,1)$
$\therefore GI=\frac{1}{3}$
$\therefore \frac{2}{3}\sqrt{{\lambda }^2-4}=\frac{1}{3}$
$\Rightarrow {\lambda }^2-4=\frac{1}{4}$
$\Rightarrow {\lambda }^2=\frac{17}{4}$
$\Rightarrow {\lambda }^2=4.2$
$\Rightarrow \left[\lambda \right]=2$ (as $\lambda >0$)