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) $\ne$ (0, 0) is

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

Answer 1

Answer: (A)

The line passing through the intersection of lines
ax + 2by = 3b = 0 and bx - 2ay - 3a = 0 is
ax + 2by + 3b + $\lambda$ (bx - 2ay - 3a) = 0
$\Rightarrow (a+b\lambda )x+(2b-2a\lambda )y+3b-3\lambda a=0$
As this line is parallel to x-axis.
$\therefore a+b\lambda =0\Rightarrow =-a/b$
$\Rightarrow ax+2by+3b-\frac{a}{b}(bx-2ay-3a)=0$
$\Rightarrow ax+2by+3b-ax+\frac{2a^2}{b}y+\frac{3a^2}{b}=0$
$y\left(2b+\frac{2a^2}{b}\right)+3b+\frac{3a^2}{b}=0$
$y\left(\frac{2b^2+2a^2}{b}\right)=-\left(\frac{3b^2+3a^2}{b}\right)$
$y=\frac{-3\left(a^2+b^2\right)}{2\left(b^2+a^2\right)}=\frac{-3}{2}$
So it is 3/2 units below x-axis.