Question

The line parallel to the x-axis and passing through the intersection of the 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.

• A. Above the x-axis at a distance of 3/2 from it
• B. Above the x-axis at a distance of 2/3 from it
• C. Below the x-axis at a distance of 3/2 from it
• D. Below the x-axis at a distance of 2/3 from it

Answer 1

Answer: (C)

Below the x-axis at a distance of 3/2 from it

Answer 2

The lines passing through the intersection of the lines $a x + 2 b y + 3 b = 0$ and $b x - 2 a y - 3 a = 0$ is

$a x + 2 b y + 3 b + \lambda ( b x - 2 a y - 3 a ) = 0$

$\Rightarrow ( a + b \lambda ) x$ $+ ( 2 b - 2 a \lambda ) y + 3 b - 3 \lambda a = 0$ …..(i)

Line (i) is parallel to x-axis,

$\therefore$ $a + b \lambda = 0 \Rightarrow \lambda = \frac { - a } { b } = 0$

Put the value of $\lambda$ in (i)

$a x + 2 b y + 3 b - \frac { a } { b } ( b x - 2 a y - 3 a ) = 0$

$y \left( 2 b + \frac { 2 a ^ { 2 } } { b } \right) + 3 b + \frac { 3 a ^ { 2 } } { b } = 0$ , $y \left( \frac { 2 b ^ { 2 } + 2 a ^ { 2 } } { b } \right) = - \left( \frac { 3 b ^ { 2 } + 3 a ^ { 2 } } { b } \right)$

$y = \frac { - 3 \left( a ^ { 2 } + b ^ { 2 } \right) } { 2 \left( b ^ { 2 } + a ^ { 2 } \right) } = \frac { - 3 } { 2 }$, $y = - \frac { 3 } { 2 }$

So, it is 3/2 unit below x-axis.