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(C) To find the intersection point,we solve the system of equations:$ax + 2by = -3b$ $(1)$$bx - 2ay = 3a$ $(2)$Multiply $(1)$ by $a$ and $(2)$ by $b$:

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Below the $x$-axis at a distance of $3/2$ from it

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(C) To find the intersection point,we solve the system of equations:$ax + 2by = -3b$ $(1)$$bx - 2ay = 3a$ $(2)$Multiply $(1)$ by $a$ and $(2)$ by $b$:$a^2x + 2aby = -3ab$$b^2x - 2aby = 3ab$Adding these equations: $(a^2 + b^2)x = 0$. Since $(a, b) 
e (0, 0)$,$a^2 + b^2 
e 0$,so $x = 0$.Substitute $x = 0$ into $(1)$: $2by = -3b$. Since $b$ could be $0$,we check the case $b=0$. If $b=0$,then $ax=0 \Rightarrow x=0$ and $-2ay=3a \Rightarrow y=-3/2$. If $b 
e 0$,$y = -3/2$.Thus,the intersection point is $(0, -3/2)$.$A$ line parallel to the $x$-axis has the form $y = k$. Since it passes through $(0, -3/2)$,the line is $y = -3/2$.This line is $3/2$ units below the $x$-axis.

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

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