Question

If the point of intersection of the lines $2ax+4ay+c=0$ and $7bx+3by-d=0$ lies in the $4^{\text{th}}$ quadrant and is equidistant from the two axes, where $a,b,c$ and $d$ are non-zero numbers, then $ad:bc$ equals to

• A. $2:3$
• B. $2:1$
• C. $1:1$
• D. $3:2$

Answer 1

The correct option is B $2:1$

Let point of intersection of $2ax+4ay+c=0$ and $7bx+3by-d=0$ be $(\alpha ,-\alpha )$.

$\Rightarrow 2a\alpha -4a\alpha +c=0$ and
$\Rightarrow 7b\alpha -3b\alpha -d=0$

$2a\alpha =c$ and $4b\alpha =d$
$\Rightarrow \frac{2a\alpha }{4b\alpha }=\frac{c}{d}$
$\Rightarrow ad:bc=2:1$