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Q.

The straight line passing through (-1,1) and remaining parallel to the line common to the pairs of lines provided by $6 x^{2} - x y - 12 y^{2} = 0$ and $15 x^{2} + 14 x y - 8 y^{2} = 0$ is

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a

$5 x - 2 y + 7 = 0$

b

$3 x + 4 y - 1 = 0$

c

$3 x - 4 y + 7 = 0$

d

$2 x - 3 y + 5 = 0$

answer is B.

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Detailed Solution

$6 x^{2} - x y - 12 y^{2} = 0 \Rightarrow (2 x - 3 y) (3 x + 4 y) 0 = 0 15 x^{2} + 14 x y - 8 y^{2} = 0 \Rightarrow (5 x - 2 y) (3 x + 4 y) = 0$
$∴$ Common line is $3 x + 4 y = 0$
$∴$ Required line is $y - 1 = \frac{- 3}{4} (x + 1) \Rightarrow 3 x + 4 y - 1 = 0$

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