The line parallel to the x-axis and passing through the intersection of the lines  where  is

The correct answer is: Below the x-axis at a distance of 3/2 from it

Here , we have to find the line of the lies.
Firstly, ax + 2by + 3b = 0 .and bx – 2ay -3a= 0 where (a,b) ≠ (0,0).
The line passing through the intersection of the lines ax+2by+3b=0 and bx−2ay−3a=0 is
ax+2by+3b+λ(bx−2ay−3a)=0...............(1)
(a+bλ)x+(2b−2aλ)y+3b−3λa=0
As the line is parallel to x−axis
a+bλ = 0
so, λ = (−a/b)
Putting λ = (−a/b) in equation (1), we get
ax+ 2by + 3b + (−a/b) (bx−2ay−3a)=0
Since it is parallel to x−axis, so coefficient of x=0. Hence we get:
⟹y (2b+ 2a²/b) + 3b+3a²/b=0
⟹ y = -3(b + a²/b)/2(b + a²/b)
⟹ y = -3/2
Therefore, it is below the x-axis at a distance of 3/2 from it.

In this question, we have to find the lines where it lies and what’s its distance from x-axis. If u=ax1+by1+c1=0 and v=ax2+by2+c2=0 are two intersecting lines, then the equation u + kv =0 represents a family of lines. Remember the family of lines