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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 equidistant from the two axes then

$\begin{array}{1 1}(A)\;2bc-3ad=0\\(B)\;2bc+3ad=0\\(C)\;3bc-2ad=0\\(D)\;3bc+2ad=0\end{array} $

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$3bc-2ad=0$
Hence (C) is the correct answer.
answered Apr 7, 2014 by sreemathi.v
 

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