Mathematics

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 $$(a,b)$$


ANSWER

Below the x-axis at a distance of $$\dfrac { 3 }{ 2 } $$ from it


SOLUTION
The line passing through the intersection of the lines $$ax + 2by + 3b = 0$$ and $$bx - 2ay - 3a = 0$$ is
$$ax + 2by + 3b+\lambda(bx - 2ay - 3a)=0...............(1)$$
$$(a+b\lambda)x+(2b-2a\lambda)y+3b-3\lambda a=0$$
As the line is parallel to $$x-axis$$
$$a+b\lambda=0$$ 
so, $$\lambda=(-\dfrac ab)$$
Putting $$\lambda=(-a/b)$$ in equation $$(1)$$, we get
$$ax + 2by + 3b+\left(-\dfrac ab\right)(bx - 2ay - 3a)=0$$
Since it is parallel to $$x-axis$$, so coefficient of $$x=0$$. Hence we get:
$$\implies y(2b+2\dfrac{a^2}{b})+3b+3\dfrac{a^2}{b}=0$$
On simplifying we get 
$$y=-\dfrac 32$$

So it is $$\dfrac32$$ units below x-axis
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