A) \[\frac{\pi }{6}\]
B) \[\frac{\pi }{4}\]
C) \[\frac{\pi }{3}\]
D) \[\frac{\pi }{2}\]
Correct Answer: D
Solution :
\[y={{x}^{2}}-5x+6\Rightarrow \frac{dy}{dx}=(2x-5)\] \[{{\left. \therefore {{m}_{1}}=\frac{dy}{dx} \right|}_{P(2,\,\,0)}}=4-5=-1\] and \[{{\left. {{m}_{2}}=\frac{dy}{dx} \right|}_{Q(3,\,\,0)}}=6-5=1\] As, \[{{m}_{1}}\times {{m}_{2}}=-1\Rightarrow \theta =\frac{\pi }{2}\]
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