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

Given lines 4ax + 2ay + c = 0---(1) and 5bx + 2by + d = 0---(2)

The point lies in Q₄ and Equidistant from axis

Then the point is ( x , -x )

Sub in (1)

 $\begin{array}{l}4ax-2ax+c=0\\ \Rightarrow 2ax+c=0\Rightarrow x=\frac{-c}{2a}\end{array}$

Sub in (2)

$\begin{array}{l}5bx-2bx+d=0\\ \Rightarrow 3bx+d=0\\ \Rightarrow x=\frac{-d}{3b}\\ \therefore \frac{-c}{2a}=\frac{-d}{3b}\\ \Rightarrow -3bc=-2ad\\ \Rightarrow 3bc-2ad=0\end{array}$