Let a, b, c and d be non-zero real numbers. I | vedclass

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

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$2ad - 3bc = 0$

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(D) Given lines are: $4ax + 2ay + c = 0$  $(i)$ $5bx + 2by + d = 0$  $(ii)$ Multiply $(i)$ by $b$ and $(ii)$ by $a$: $4abx + 2aby + bc = 0$ $5abx + 2aby + ad = 0$ Subtracting the two equations: $-abx + bc - ad = 0 \Rightarrow x = \frac{bc - ad}{ab}$ Substitute $x$ into $(i)$: $4a(\frac{bc - ad}{ab}) + 2ay + c = 0$ $4(\frac{bc - ad}{b}) + 2ay + c = 0$ $2ay = -c - \frac{4bc - 4ad}{b} = \frac{-bc - 4bc + 4ad}{b} = \frac{4ad - 5bc}{b}$ $y = \frac{4ad - 5bc}{2ab}$ Since the point lies in the $4^{	ext{th}}$ quadrant and is equidistant from the axes,$x = -y$ and $x > 0, y $\frac{bc - ad}{ab} = -(\frac{4ad - 5bc}{2ab})$ $2(bc - ad) = -(4ad - 5bc)$ $2bc - 2ad = -4ad + 5bc$ $2ad - 3bc = 0$

Let $a, b, c$ and $d$ be non-zero real numbers. If the point of intersection of the lines $4ax + 2ay + c = 0$ and $5bx + 2by + d = 0$ lies in the $4^{\text{th}}$ quadrant and is equidistant from the two axes,then:

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