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 $7$ from it is

3 x - 4 y = - 35

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5 x - 2 y = 7

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3 x + 4 y = 35

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2 x - 3 y = 7

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Solution

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

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