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 $(a,b)$

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

Answer 1

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

The line passing through the intersection of the lines $ax+2by+3b=0$ and $bx-2ay-3a=0$ is

$ax+2by+3b+\lambda (bx-2ay-3a)=\mathrm{0...............}\left(1\right)$

$(a+b\lambda )x+(2b-2a\lambda )y+3b-3\lambda a=0$

As the line is parallel to $x-axis$

$a+b\lambda =0$

so, $\lambda =(-\frac{a}{b})$

Putting $\lambda =(-a/b)$ in equation $\left(1\right)$, we get

$ax+2by+3b+(-\frac{a}{b})(bx-2ay-3a)=0$

Since it is parallel to $x-axis$, so coefficient of $x=0$. Hence we get:

$⟹y(2b+2\frac{a^2}{b})+3b+3\frac{a^2}{b}=0$

On simplifying we get

$y=-\frac{3}{2}$

So it is $\frac{3}{2}$ units below x-axis