JEE MAIN - Mathematics (2005 - No. 45)
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 :
below the $$x$$ - axis at a distance of $${3 \over 2}$$ from it
below the $$x$$ - axis at a distance of $${2 \over 3}$$ from it
above the $$x$$ - axis at a distance of $${3 \over 2}$$ from it
above the $$x$$ - axis at a distance of $${2 \over 3}$$ from it
Explanation
The line passing through the intersection of lines
$$ax + 2by = 3b = 0$$
and $$bx - 2ay - 3a = 0$$ is
$$ax + 2by + 3b + \lambda \left( {bx - 2ay - 3a} \right) = 0$$
$$ \Rightarrow \left( {a + b\lambda } \right)x + \left( {2b - 2a\lambda } \right)y + 3b - 3\lambda a = 0$$
As this line is parallel to $$x$$-axis.
$$\therefore$$ $$a + b\lambda = 0 \Rightarrow \lambda = - a/b$$
$$ \Rightarrow ax + 2by + 3b - {a \over b}\left( {bx - 2ay - 3a} \right) = 0$$
$$ \Rightarrow ax + 2by + 3b - ax + {{2{a^2}} \over b}y + {{3{a^2}} \over b} = 0$$
$$y\left( {2b + {{2{a^2}} \over b}} \right) + 3b + {{3{a^2}} \over b} = 0$$
$$y\left( {{{2{b^2} + 2{a^2}} \over b}} \right) = - \left( {{{3{b^2} + 3{a^2}} \over b}} \right)$$
$$y = {{ - 3\left( {{a^2} + {b^2}} \right)} \over {2\left( {{b^2} + {a^2}} \right)}} = {{ - 3} \over 2}$$
So it is $$3/2$$ units below $$x$$-axis.
$$ax + 2by = 3b = 0$$
and $$bx - 2ay - 3a = 0$$ is
$$ax + 2by + 3b + \lambda \left( {bx - 2ay - 3a} \right) = 0$$
$$ \Rightarrow \left( {a + b\lambda } \right)x + \left( {2b - 2a\lambda } \right)y + 3b - 3\lambda a = 0$$
As this line is parallel to $$x$$-axis.
$$\therefore$$ $$a + b\lambda = 0 \Rightarrow \lambda = - a/b$$
$$ \Rightarrow ax + 2by + 3b - {a \over b}\left( {bx - 2ay - 3a} \right) = 0$$
$$ \Rightarrow ax + 2by + 3b - ax + {{2{a^2}} \over b}y + {{3{a^2}} \over b} = 0$$
$$y\left( {2b + {{2{a^2}} \over b}} \right) + 3b + {{3{a^2}} \over b} = 0$$
$$y\left( {{{2{b^2} + 2{a^2}} \over b}} \right) = - \left( {{{3{b^2} + 3{a^2}} \over b}} \right)$$
$$y = {{ - 3\left( {{a^2} + {b^2}} \right)} \over {2\left( {{b^2} + {a^2}} \right)}} = {{ - 3} \over 2}$$
So it is $$3/2$$ units below $$x$$-axis.
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