Discussion Forum

Let a, b, c and d be non-zero numbers. If the point of intersection of lines 4ax+2ay+c=0 and 5bx+2by+d=0 lies in the fourth quadrant and is equidistant from the two axes, then

Answer:

Explanation:

 Let coordinate of the intersction point in fourth quadrant be $(\alpha,-\alpha)$

 Since, $(\alpha,-\alpha)$   lies on both lines
        

                    4ax+2ay+c=0

 and 5bx+2by+d=0

$\therefore$      $4a\alpha-2a\alpha+c=0$

     $\Rightarrow$    $ \alpha=\frac{-c}{2a}$               ......(i)

  and                   $5b\alpha-2b\alpha+d=0$

   $\Rightarrow$  $\alpha=\frac{-d}{3b}$        .........(ii)

From eqs.(i) and (ii)  , we get

     $\frac{-c}{2a}=\frac{-d}{3b}$

$\Rightarrow $  3bc=2ad

$\Rightarrow  $   2ad-3bc=0