Lines $(1 + λ) x + (4 - λ) y + (2 + λ) = 0$ and $(4 - λ) x + (1 + λ) y + (6 - 3 λ) = 0$ are concurrent at points A and B respectively and intersect at C, then locus of centroid of $Δ A B C$ is

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${(x + \frac{3}{2})}^{2} + {(y + \frac{7}{10})}^{2} = \frac{17}{450}$

${(x + \frac{3}{2})}^{2} + {(y + \frac{7}{10})}^{2} = \frac{17}{225}$

${(x - \frac{3}{2})}^{2} + {(y - \frac{7}{10})}^{2} = \frac{17}{50}$

${(x + \frac{3}{2})}^{2} + {(y + \frac{7}{10})}^{2} = \frac{17}{50}$

answer is C.

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Detailed Solution

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 $\frac{1}{3} (P A)$ 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., $(\frac{- 9}{5}, - \frac{6}{5})$
$\Rightarrow P = (\frac{- 3}{2}, \frac{- 7}{10})$ therefore locus is ${(x + \frac{3}{2})}^{2} + {(y + \frac{7}{10})}^{2} = \frac{17}{50}$