The centroid of the triangle formed by the lines $2 x - 3 y - 4 = 0 a n d 12 x^{2} - 20 x y + 7 y^{2} = 0$

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$(\frac{5}{3}, \frac{5}{3})$

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

$(\frac{11}{3}, \frac{11}{3})$

$(\frac{1}{3}, \frac{1}{3})$

answer is A.

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Detailed Solution

$\begin{array}{l} G i v e n E q u a t i o n 12 x^{2} - 20 x y + 7 y^{2} = 0 \\ \Rightarrow 12 x^{2} - 6 x y - 14 x y + 7 y^{2} = 0 \\ \Rightarrow 6 x (2 x - y) - 7 y (2 x - y) = 0 \\ \Rightarrow (2 x - y) (6 x - 7 y) = 0 \\ \Rightarrow 2 x - y = 0 - - (1), 6 x - 7 y = 0 - - - (2) a n d 2 x - 3 y - 4 = 0 - - (3) \\ P . I . o f (1) & (2) i s (0, 0) \\ s o l v i n g (1) & (3) \\ x y 1 \\ \begin{matrix} - 1 & 0 & 2 & - 1 \\ - 3 & - 4 & 2 & - 3 \end{matrix} \\ \Rightarrow \frac{x}{4 - 0} = \frac{y}{0 + 8} = \frac{1}{- 6 + 2} \\ \Rightarrow \frac{x}{4} = \frac{y}{8} = \frac{1}{- 4} \\ (x, y) = (- 1, - 2) \end{array}$

$\begin{array}{l} S o l v i n g (2) & (3) \\ x y 1 \\ \begin{matrix} - 7 & 0 & 6 & - 7 \\ - 3 & - 4 & 2 & - 3 \end{matrix} \\ \Rightarrow \frac{x}{28 - 0} = \frac{y}{0 + 24} = \frac{1}{- 18 + 14} \\ \Rightarrow \frac{x}{28} = \frac{y}{24} = \frac{1}{- 4} \\ (x, y) = (- 7, - 6) \\ O (0, 0) A (- 1, - 2) B (- 7, - 6) \\ C e n t r o i d = (\frac{0 - 1 - 7}{3}, \frac{0 - 2 - 6}{3}) \Rightarrow (\frac{- 8}{3}, \frac{- 8}{3}) \end{array}$