Let a, b, c and d be non-zero numbers. If the point of intersection of the lines 4ax+2ay+c=0 and 5bx+2by+d=0 lies in the fourth quadrant and is equdistant from the two axes, then
A
2bc−3d=0
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B
2bc+3ad=0
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C
2ad−3bc=0
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D
3bc+2ad=0
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Solution
The correct option is C2ad−3bc=0 Let coordinate of the intersection point in fourth quadrant be (α,−α).
Since (α,−α) lies on both lines 4ax+2ay+c=0 and 5bx+2by+d=0 ∴4aα−2aα+c=0⇒α=−c2a …..(i)
And 5bα−2bα+d=0⇒α=−d3b ….(ii)
From Eqs. (i) and (ii), we get −c2a=−d3b⇒3bc=2ad ⇒2ad−3bc=0