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