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

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a

3x−4y=−35

b

5x−2y=7

c

3x+4y=35

d

2x−3y=7

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detailed solution

Correct option is C

We have 6x2−xy−12y2=0 or (2x−3y)(3x+4y)=0-----(1) and 15x2+14xy−8y2=0 or (5x−2y)(3x+4y)=0---(2) Equation of the line common to (1) and (2) is 3x+4y=0---(3)Equation of any line parallel to (3) is 3x+4y=kSince its distance from (3) is 7 , we have c2-c1a2+b2=k32+42=7 or k=±35

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