Lines $(1+\lambda )x+(4-\lambda )y+(2+\lambda )=0\mathrm{and}(4-\lambda )x-(1+\lambda )y+(6-3\lambda )=0$ are concurrent at points A and B respectively and intersect at C, then locus of centroid of $\Delta ABC$ is

As lines are perpendicular to each other ‘C’ moves on a circle with AB as diameter. Now P, mid–point of AB (which is fixed) when joined with C is median.

$\Rightarrow$Centroid is moving at a constant distance $\frac{1}{3}\left(PA\right)$ from P

$\Rightarrow$Locus is a circle

A is point of intersection of $x+4y+2=0$ and $x–y+1=0 i.e.,(-\frac{6}{5},-\frac{1}{5})$

B is point of intersection of $4x–y+6=0$ and $x+y+3=0 i.e.,(-\frac{9}{5},-\frac{6}{5})$

$\Rightarrow P=(-\frac{3}{2},\frac{-7}{10})thereforelocusis{(x+\frac{3}{2})}^2+{(y+\frac{7}{10})}^2=\frac{17}{450}$