A) \[{{h}^{2}}=a+b\]
B) \[a+b=0\]
C) \[{{h}^{2}}=ab\]
D) \[h=0\]
Correct Answer: B
Solution :
We know, if the lines are perpendicular to each other, then \[\theta ={{90}^{o}}\] \[\Rightarrow \] \[\tan \theta =\frac{2\sqrt{{{h}^{2}}-ab}}{a+b}\] \[\Rightarrow \] \[\tan {{90}^{o}}=\frac{2\sqrt{{{h}^{2}}-ab}}{a+b}\] \[\Rightarrow \] \[\frac{1}{0}=\frac{2\sqrt{{{h}^{2}}-ab}}{a+b}\] \[\Rightarrow \] \[a+b=0\]
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