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The line parallel to the $x$-axis and passing through the intersection of the lines ax $+$ $2 b y+3 b=0$ and $b x-2 a y-3 a=0$, where $(a, b) \neq(0,0)$ is
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The correct answer is:
below the $x$-axis at a distance of $\frac{3}{2}$ from it
below the $x$-axis at a distance of $\frac{3}{2}$ from it
$$
\begin{aligned}
& a x+2 b y+3 b+\lambda(b x-2 a y-3 a)=0 \\
& \Rightarrow(a+b \lambda) x+(2 b-2 a \lambda) y+3 b-3 \lambda a=0 \\
& a+b \lambda=0 \Rightarrow \lambda=-a / b \\
& \Rightarrow a x+2 b y+3 b-\frac{a}{b}(b x-2 a y-3 a)=0 \\
& \Rightarrow a x+2 b y+3 b-a x+\frac{2 a^2}{b} y+\frac{3 a^2}{b}=0 \\
& y\left(2 b+\frac{2 a^2}{b}\right)+3 b+\frac{3 a^2}{b}=0 \\
& y\left(\frac{2 b^2+2 a^2}{b}\right)=-\left(\frac{3 b^2+3 a^2}{b}\right) \\
& y=\frac{-3\left(a^2+b^2\right)}{2\left(b^2+a^2\right)}=\frac{-3}{2} \\
&
\end{aligned}
$$
$y=-\frac{3}{2}$ so it is $3 / 2$ units below $x$-axis.
\begin{aligned}
& a x+2 b y+3 b+\lambda(b x-2 a y-3 a)=0 \\
& \Rightarrow(a+b \lambda) x+(2 b-2 a \lambda) y+3 b-3 \lambda a=0 \\
& a+b \lambda=0 \Rightarrow \lambda=-a / b \\
& \Rightarrow a x+2 b y+3 b-\frac{a}{b}(b x-2 a y-3 a)=0 \\
& \Rightarrow a x+2 b y+3 b-a x+\frac{2 a^2}{b} y+\frac{3 a^2}{b}=0 \\
& y\left(2 b+\frac{2 a^2}{b}\right)+3 b+\frac{3 a^2}{b}=0 \\
& y\left(\frac{2 b^2+2 a^2}{b}\right)=-\left(\frac{3 b^2+3 a^2}{b}\right) \\
& y=\frac{-3\left(a^2+b^2\right)}{2\left(b^2+a^2\right)}=\frac{-3}{2} \\
&
\end{aligned}
$$
$y=-\frac{3}{2}$ so it is $3 / 2$ units below $x$-axis.
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