Straight Line Question 158
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 (a, b) $ \ne $ (0, 0) is
Options:
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 $ \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{2a^{2}}{b}y+\frac{3a^{2}}{b}=0 $ $ \Rightarrow y( 2y+\frac{2a^{2}}{b} )+\frac{3a^{2}}{b}+3b=0 $ $ \Rightarrow y( \frac{2b^{2}+2a^{2}}{b} )=-( \frac{3b^{2}+3a^{2}}{b} ) $ $ \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.Show Answer
Answer:
Solution:
BETA
