Equation of a line which is parallel to the line common to the pair of lines given by 6x2−xy−12y2=0 and 15x2+14xy−8y2=0 and at a distance of 7 units from it, is
A
3x+4y=35
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B
3x+4y+35=0
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C
2x−3y=35
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D
2x−3y+35=0
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Solution
The correct option is B3x+4y+35=0 We have, 6x2−xy−12y2=0 ⇒(2x−3y)(3x+4y)=0⋯(1)
and 15x2+14xy−8y2=0 ⇒(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=k
Since, its distance from (3) is 7 units,
so ∣∣
∣∣k√32+42∣∣
∣∣=7 ⇒k=±35