Question

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

• A. $2bc - 3ad = 0$
• B. $2bc + 3ad = 0$
• C. $3bc - 2ad= 0$
• D. $3bc + 2ad = 0$

Answer 1

Answer: (C)

$3bc - 2ad= 0$

Answer 2

Let $(\alpha,-\alpha)$ be the point of intersection
$\therefore 4 a \alpha-2 a \alpha+ c =0$
$\Rightarrow \alpha=-\frac{c}{2a}$
and $5b \alpha-2 b \alpha+ d =0$
$\Rightarrow \alpha=-\frac{d}{3 b}$
$\Rightarrow 3 b c=2 a d$
$\Rightarrow 3 b c-2 a d=0$
:
The point of intersection will be
$\frac{x}{2 a d-2 b c}=\frac{-y}{4 a d-5 b c}=\frac{1}{8 a b-10 a b}$
$\Rightarrow x=\frac{2(a d-b c)}{-2 a b}$
$\Rightarrow y=\frac{5 b c-4 a d}{-2 a b}$
$\because$ Point of intersection is in fourth quadrant so x is positive and y is negative
Also distance from axes is same
So $x = - y$ $(\because$ distance from x-axis is - y as y is negative)

$\frac{2(a d-b c)}{-2 a b}=\frac{-(5 b c-4 a d)}{-2 a b}$
$2 a d-2 b c=-5 b c+4 a d$
$\Rightarrow 3 b c-2 a d=0\,\,\,\,\,\,\dots(i)$