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The orthocentre of a triangle formed by the line $x + y = 3$ and pair of lines $6 x^{2} - 5 x y - 6 y^{2} + x + 5 y - 1 = 0$ is

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$(\frac{13}{169}, \frac{- 65}{169})$

$(\frac{- 13}{169}, \frac{- 65}{169})$

$(\frac{- 13}{169}, \frac{65}{169})$

$(\frac{13}{169}, \frac{65}{169})$

answer is A.

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Detailed Solution

$a = 2, 2 h = - 5, b = - 6, 2 g = 1, 2 f = 5, c = - 1$

$a + b = 6 - 6 = 0$ , triangle is right angled orhtocentre I. E point of intersection of pair of lines

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