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Q.
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) $\neq$ (0, 0) is
Straight Lines
Solution:
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.