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Q.

The product of the perpendicular distances drawn from the origin to the pair of straight lines $6 x^{2} - 5 x y - 6 y^{2} + x + 5 y - 1 = 0$ is

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a

1

b

13

c

$\frac{1}{12}$

d

$\frac{1}{13}$

answer is C.

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Detailed Solution

Given $6 x^{2} - 5 x y - 6 y^{2} + x + 5 y - 1 = 0$
Perpendicular distance
$\frac{| c |}{\sqrt{{(a - b)}^{2} + 4 h^{2}}} = \frac{| - 1 |}{\sqrt{{(6 + 6)}^{2} + 25}} = \frac{1}{13}$

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