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 C
$3 x + 4 y = - 35$

$6 x^{2} - x y - 12 y^{2} = 0$
$\Rightarrow$ (2x - 3y)(3x + 4y) = 0 (i)
and $15 x^{2}$ + 14xy - $8 y^{2}$ = 0
$\Rightarrow$ (5x - 2y) (3x + 4y) = 0 (ii)
Equation of the line common to (i) and (ii) is
3x + 4y = 0 (iii)
Equation of any line parallel to (ii) is
3x + 4y = k
Since its distance from (iii) is 7
$∣ ∣ ∣ \frac{k}{\sqrt{3^{2} + 4^{2}}} ∣ ∣ ∣ = 7 \Rightarrow k = \pm (7 \times 5) = \pm 35$

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