Q38 / 90 · JEE Main 2012 (26 May Online) · Mathematics
The line parallel to $x$-axis and passing through the point of intersection of lines $a x+2 b y+3 b=0$ and $b x-2 a y-3 a=0$, where $(a, b) \neq(0,0)$ is
- Aabove $x$-axis at a distance $2 / 3$ from it
- Babove $x$-axis at a distance $3 / 2$ from it
- Correct answerCbelow $x$-axis at a distance $3 / 2$ from it
- Dbelow $x$-axis at a distance $2 / 3$ from it
Solution
Given lines are
$$
a x+2 b y+3 b=0 \text { and }
$$
$$
b x-2 a y-3 a=0
$$
Since, required line is $\|$ to $x$-axis
$$
\therefore x=0
$$
We put $x=0$ in given equation, we get
$$
2 b y=-3 b \Rightarrow y=-\frac{3}{2}
$$
This shows that the required line is below $x$-axis at a distance of $\frac{3}{2}$ from it.
Hence, the correct answer is option: 'C'.