1. Given lines

The two lines are: $4ax+2ay+c=0$ $5bx+2by+d=0$

We need the point of intersection of these lines.

2. Find the intersection point

From $4ax+2ay+c=0$ we get $2a(2x+y)=-c$ $2x+y=-\frac{c}{2a}$

From $5bx+2by+d=0$ we get $b(5x+2y)=-d$ $5x+2y=-\frac{d}{b}$

So we solve the system: $2x+y=-\frac{c}{2a} \quad ...(1)$ $5x+2y=-\frac{d}{b} \quad ...(2)$

Multiply (1) by 2: $4x+2y=-\frac{c}{a}$

Subtract this from (2): $(5x+2y)-(4x+2y)=-\frac{d}{b}+\frac{c}{a}$ $x=\frac{c}{a}-\frac{d}{b}$

Taking LCM: $x=\frac{bc-ad}{ab}$

Now from (1): $y=-\frac{c}{2a}-2x$ $y=-\frac{c}{2a}-2\left(\frac{bc-ad}{ab}\right)$

Using common denominator $2ab$: $y=\frac{-bc-4bc+4ad}{2ab}$ $y=\frac{4ad-5bc}{2ab}$

Thus intersection point is $\left(\frac{bc-ad}{ab},\; \frac{4ad-5bc}{2ab}\right)$

3. Use the condition: point lies in fourth quadrant and is equidistant from axes

A point equidistant from the two axes satisfies $|x|=|y|$

Since the point is in the fourth quadrant, we have $x>0,\quad y<0$ Therefore, $y=-x$

So, $\frac{4ad-5bc}{2ab}=-\frac{bc-ad}{ab}$

Multiply both sides by $2ab$: $4ad-5bc=-2(bc-ad)$ $4ad-5bc=-2bc+2ad$

Bring all terms to one side: $4ad-5bc-2ad+2bc=0$ $2ad-3bc=0$

Hence, $3bc-2ad=0$

4. Match with options

This is exactly: $\boxed{3bc-2ad=0}$ So the correct option is:

A

5. Comparison with stored answer

Stored correct answer: A

Our derived answer: A

They agree.