Let a, b, c and d be non-zero numbers. If the | vedclass

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(A) Let the point of intersection in the fourth quadrant be $(\alpha, -\alpha)$,where $\alpha > 0$.Since this point lies on both lines $4ax + 2ay + c

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$3bc - 2ad = 0$

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(A) Let the point of intersection in the fourth quadrant be $(\alpha, -\alpha)$,where $\alpha > 0$.Since this point lies on both lines $4ax + 2ay + c = 0$ and $5bx + 2by + d = 0$,we substitute the coordinates:For the first line: $4a(\alpha) + 2a(-\alpha) + c = 0$ $\Rightarrow 2a\alpha + c = 0$ $\Rightarrow \alpha = -\frac{c}{2a}$.For the second line: $5b(\alpha) + 2b(-\alpha) + d = 0$ $\Rightarrow 3b\alpha + d = 0$ $\Rightarrow \alpha = -\frac{d}{3b}$.Equating the two expressions for $\alpha$:$-\frac{c}{2a} = -\frac{d}{3b}$.Cross-multiplying gives $3bc = 2ad$,which can be written as $3bc - 2ad = 0$.

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 axes,then:

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