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 $(a,b)$

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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Single Correct Medium Published on 09th 09, 2020
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