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
Correct Answer: C
Solution :
The lines passing through the intersection of the lines \[ax+2by+3b=0\] and \[bx-2ay-3a=0\] is \[ax+2by+3b+\lambda (bx-2ay-3a)=0\] \[\Rightarrow (a+b\lambda )x\]\[+(2b-2a\lambda )y+3b-3\lambda a=0\] ?..(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) \[ax+2by+3b-\frac{a}{b}(bx-2ay-3a)=0\] \[y\left( 2b+\frac{2{{a}^{2}}}{b} \right)+3b+\frac{3{{a}^{2}}}{b}=0\], \[y\left( \frac{2{{b}^{2}}+2{{a}^{2}}}{b} \right)=-\left( \frac{3{{b}^{2}}+3{{a}^{2}}}{b} \right)\] \[y=\frac{-3({{a}^{2}}+{{b}^{2}})}{2({{b}^{2}}+{{a}^{2}})}=\frac{-3}{2}\], \[y=-\frac{3}{2}\] So, it is 3/2 unit below x-axis.
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