Question

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

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

Answer 1

The correct option is D

Below the$X-$axis at a distance of $\frac{3}{2}$

Find the equation of the line passing through the point of intersection of the other lines

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

$\Rightarrow ax+2by+3b+\lambda (bx-2ay-3a)=0$

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

$\Rightarrow y=\left[\frac{-(a+b\lambda )}{(2b–2a\lambda )}\right]x-\frac{(3b–3a\lambda )}{(2b–2a\lambda )}.......\left(1\right)$

Here, the slope, $m=\left[\frac{-(a+b\lambda )}{\left(2b-2a\lambda \right)}\right]$

Since, it is parallel to $X$-axis, so slope $m=0$

$\Rightarrow \left[\frac{-(a+b\lambda )}{\left(2b-2a\lambda \right)}\right]=0$

$\Rightarrow \lambda =\frac{-a}{b}$

Putting the value of $\lambda$in $\left(1\right)$, we get

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

It is $\frac{3}{2}\mathrm{units}$ below the $X$ axis

Hence, the correct answer is Option $\left(\mathrm{D}\right)$.