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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} $

1 Answer

Since they are equidistant and line in the forth quadrant, let the coordinate be $(h, -h)$
$3ah - 2ah +c = 0$
$5bh - 2bh +d = 0$
solving both the equation we get
$\begin{align*} \frac{-c}{2a} = -\frac{d}{3b} \end{align*}$
$\implies 3bc - 2ad = 0$
answered Apr 7, 2014 by sreemathi.v
edited Nov 7, 2017 by priyanka.c

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