Let a,b,c,d be non-zero numbers. If point of intersection of 6ax+2ay+c=0 and 5bx+3by+d=0 lies in second quadrant such that its distance from x-axis is twice its distance from y-axis then
A
7bc+10ad=0
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B
bc+2ad+=0
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C
7ad+10bc=0
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D
bc−2ad=0
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Solution
The correct option is Abc+2ad+=0 Let the coordinates of the intersection point in the second quadrant be (−α,2α) lies on both lines
6ax+2ay+c=0 and 5bx+3by+d=0
The coordinates of the intersection point are represented as (−α,2α) because it is given that the distance from x-axis twice the distance from the y-axis
These coordinates should satisfy the equations of the intersecting lines