Let $A = (2, 5)$ and $B = (4, - 1)$ are two vertices of a $∆ ABC .$ Third vertex $C$ moves along $L = 9 x + 7 y + 4 = 0 .$ The locus of the centroid of triangle $ABC$ is the line

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$27 x + 21 y - 77 = 0$

$9 x - 7 y - 22 = 0$

$9 x + 7 y + 22 = 0$

$27 x + 21 y - 78 = 0$

answer is D.

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Detailed Solution

Given two vertices of triangle are A=(2, 5), B=(4, -1)

Let $C = (x_{3}, y_{3})$ be third vertex lies on the line $9 x + 7 y + 4 = 0$

$\Rightarrow 9 x_{3} + 7 y_{3} + 4 = 0 - (1)$

Let (h, k) be centriod of $∆^{le} ABC$ then

$(h, k) = (\frac{2 + 4 + x_{3}}{3}, \frac{5 - 1 + y_{3}}{3}) \Rightarrow (h, k) = (\frac{6 + x_{3}}{3}, \frac{4 + y_{3}}{3}) \Rightarrow h = \frac{6 + x_{3}}{3}, k = \frac{4 + y_{3}}{3} \Rightarrow x_{3} = 3 h - 6, y_{3} = 3 k - 4 from (1) \Rightarrow 9 (3 h - 6) + 7 (3 k - 4) + 4 = 0 \Rightarrow 27 h - 54 + 21 k - 28 + 4 = 0 \Rightarrow 27 h + 21 k - 78 = 0 ∴ locus of (h, k) is 27 x + 21 y - 78 = 0$