Question

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

• A. above the $x-$axis at a distance of $\frac{3}{2}$ units from it
• B. above the $x-$axis at a distance of $\frac{2}{3}$ units from it
• C. below the $x-$axis at a distance of $\frac{3}{2}$ units from it
• D. below the $x-$axis at a distance of $\frac{2}{3}$ units from it

Answer 1

The correct option is C below the $x-$axis at a distance of $\frac{3}{2}$ units from it

Solving $ax+2by+3b=0.......\left(i\right)$ and $bx-2ay-3a=\mathrm{0........}\left(ii\right)$

We get $y=\frac{-3}{2}$
Line parallel to x-axis is $y=constant$
Thus the required line is $y=\frac{-3}{2}$
Hence the line is $\frac{3}{2}$ units below x axis.