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The centroid of the triangle formed by the pair of straight lines $12 x^{2} - 20 x y + 7 y^{2} = 0$ and the line $2 x - 3 y + 4 = 0$ is

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$(- \frac{7}{3}, \frac{7}{3})$

$(- \frac{8}{3}, \frac{8}{3})$

$(\frac{8}{3}, \frac{8}{3})$

$(\frac{4}{3}, \frac{4}{3})$

answer is C.

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Detailed Solution

$a = 12, 2 h = - 10, b = 7, l = 2, m = - 3, n = 4$

$\frac{α}{b l - h m} = \frac{β}{a m - h l}$

$= \frac{- 2 n}{3 (a m^{2} - 2 h l m + b l^{2})} \Rightarrow (α, β) = (\frac{8}{3}, \frac{8}{3})$

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