Equation of a line which is parallel to the line common to the pair of lines given by $6 x^{2} - x y - 12 y^{2} = 0$ and $15 x^{2} + 14 x y - 8 y^{2} = 0$ and at a distance of $7$ units from it, is

3 x + 4 y = 35

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3 x + 4 y + 35 = 0

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2 x - 3 y = 35

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2 x - 3 y + 35 = 0

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Solution

The correct option is B $3 x + 4 y + 35 = 0$
We have, $6 x^{2} - x y - 12 y^{2} = 0$
$\Rightarrow (2 x - 3 y) (3 x + 4 y) = 0 \dots (1)$
and $15 x^{2} + 14 x y - 8 y^{2} = 0$
$\Rightarrow (5 x - 2 y) (3 x + 4 y) = 0 \dots (2)$
Equation of the line common to $(1)$ and $(2)$ is $3 x + 4 y = 0 \dots (3)$
Equation of any line parallel to $(3)$ is $3 x + 4 y = k$
Since, its distance from $(3)$ is $7$ units,
so $∣ ∣ ∣ ∣ \frac{k}{\sqrt{3^{2} + 4^{2}}} ∣ ∣ ∣ ∣ = 7$
$\Rightarrow k = \pm 35$

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