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(A) Given lines are $4ax + 2ay + c = 0$ $(i)$ and $5bx + 2by + d = 0$ (ii).Multiply $(i)$ by $b$ and (ii) by $a$:$4abx + 2aby + bc = 0$ (iii)$5abx + 2

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$3bc - 2ad = 0$

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(A) Given lines are $4ax + 2ay + c = 0$ $(i)$ and $5bx + 2by + d = 0$ (ii).Multiply $(i)$ by $b$ and (ii) by $a$:$4abx + 2aby + bc = 0$ (iii)$5abx + 2aby + ad = 0$ (iv)Subtracting (iii) from (iv):$(5ab - 4ab)x + (ad - bc) = 0$ $\Rightarrow abx = bc - ad$ $\Rightarrow x = \frac{bc - ad}{ab}$.Substitute $x$ into (ii):$5b(\frac{bc - ad}{ab}) + 2by + d = 0 \Rightarrow \frac{5(bc - ad)}{a} + 2by + d = 0$.$2by = -d - \frac{5bc - 5ad}{a} = \frac{-ad - 5bc + 5ad}{a} = \frac{4ad - 5bc}{a}$.$y = \frac{4ad - 5bc}{2ab}$.The point $P(x, y)$ is in the fourth quadrant,so $x > 0$ and $y $\frac{bc - ad}{ab} = -(\frac{4ad - 5bc}{2ab}) \Rightarrow \frac{bc - ad}{ab} = \frac{5bc - 4ad}{2ab}$.$2(bc - ad) = 5bc - 4ad \Rightarrow 2bc - 2ad = 5bc - 4ad$.$2ad = 3bc \Rightarrow 3bc - 2ad = 0$.

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 coordinate axes,then

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