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Question

(A) The family of lines passing through the intersection of $ax + 2by + 3b = 0$ and $bx - 2ay - 3a = 0$ is given by $(ax + 2by + 3b) + \lambda(bx - 2a

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At a distance of $3/2$ below the $x$-axis

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(A) The family of lines passing through the intersection of $ax + 2by + 3b = 0$ and $bx - 2ay - 3a = 0$ is given by $(ax + 2by + 3b) + \lambda(bx - 2ay - 3a) = 0$.Rearranging the terms,we get $(a + \lambda b)x + (2b - 2a\lambda)y + (3b - 3a\lambda) = 0$.Since the line is parallel to the $x$-axis,the coefficient of $x$ must be zero.Therefore,$a + \lambda b = 0$,which gives $\lambda = -a/b$.Substituting $\lambda = -a/b$ into the equation:$(ax + 2by + 3b) - \frac{a}{b}(bx - 2ay - 3a) = 0$$ax + 2by + 3b - ax + \frac{2a^2}{b}y + \frac{3a^2}{b} = 0$$y(2b + \frac{2a^2}{b}) + (3b + \frac{3a^2}{b}) = 0$$y(\frac{2b^2 + 2a^2}{b}) = -(\frac{3b^2 + 3a^2}{b})$$y = -\frac{3(a^2 + b^2)}{2(a^2 + b^2)} = -\frac{3}{2}$.Thus,the line is $y = -3/2$,which is at a distance of $3/2$ below the $x$-axis.

The line passing through the intersection of the lines $ax + 2by + 3b = 0$ and $bx - 2ay - 3a = 0$ (where $(a, b) \neq (0, 0)$) and parallel to the $x$-axis is:

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