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
$$3bc - 2ad = 0$$
$$3bc + 2ad = 0$$
$$2bc - 3ad = 0$$
$$2bc + 3ad = 0$$
Since the point of intersection lies on fourth quadrant and equidistant from the two axes,
i.e., let the point be (k, $$-$$k) and this point satisfies the two equations of the given lines.
$$\therefore$$ 4ak $$-$$ 2ak + c = 0 ......... (1)