A) 1
B) 2
C) 3
D) 4
Correct Answer: B
Solution :
It is given that the lines \[ax+by+p=0\]and \[x\cos \alpha +y\sin \alpha =p\] are inclined at an angle \[\frac{\pi }{4}\]. Therefore \[\tan \frac{\pi }{4}=\frac{-\frac{a}{b}+\frac{\cos \alpha }{\sin \alpha }}{1+\frac{a\cos \alpha }{b\sin \alpha }}\] Þ \[a\cos \alpha +b\sin \alpha =-a\sin \alpha +b\cos \alpha \] .....(i) It is given that the lines \[ax+by+p=0\], \[x\cos \alpha +y\sin \alpha -p=0\] and \[x\sin \alpha -y\cos \alpha =0\] are concurrent. \ \[\left| \,\begin{matrix} a & b & p \\ \cos \alpha & \sin \alpha & -p \\ \sin \alpha & -\cos \alpha & 0 \\ \end{matrix}\, \right|=0\] Þ \[-ap\cos \alpha -bp\sin \alpha -p=0\Rightarrow -a\cos \alpha -b\sin \alpha =1\] Þ \[a\cos \alpha +b\sin \alpha =-1\] ......(ii) From (i) and (ii), \[-a\sin \alpha +b\cos \alpha =-1\] From (ii) and (iii), \[{{(a\cos \alpha +b\sin \alpha )}^{2}}+{{(-a\sin \alpha +b\cos \alpha )}^{2}}=2\] Þ \[{{a}^{2}}+{{b}^{2}}=2\].
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