A) below the x-axis at a distance of 3/2 from it.
B) below the x-axis at a distance of 2/3 from it.
C) above the x-axis at a distance of 2/3 form it.
D) above the x-axis at a distance of 2/3 from it.
Correct Answer: A
Solution :
[a] Line through point of intersection of given two lines is \[ax+2by+3b+\lambda (bx-2ay-3a)=0\] \[\Rightarrow (a+b\lambda )x+(2b-2a\lambda )y+3b-3\lambda a=0\] Since line is parallel to x-axis, \[a+b\lambda =0\] \[\Rightarrow \lambda =-a/b\] \[\therefore ax+2by+3b=\frac{a}{b}(bx-2ay-3a)=0\] \[\Rightarrow ax+2by+3b-ax+\frac{2{{a}^{2}}}{b}y+\frac{3{{a}^{2}}}{b}=0\] \[\Rightarrow y\left( 2y+\frac{2{{a}^{2}}}{b} \right)+\frac{3{{a}^{2}}}{b}+3b=0\] \[\Rightarrow y\left( \frac{2{{b}^{2}}+2{{a}^{2}}}{b} \right)=-\left( \frac{3{{b}^{2}}+3{{a}^{2}}}{b} \right)\] \[\Rightarrow y=\frac{-3({{a}^{2}}+{{b}^{2}})}{2({{b}^{2}}+{{a}^{2}})}=\frac{-3}{2}\] So it is 3/2 units below the x-axis.
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