• # question_answer If straight lines $ax+by+p=0$ and $x\cos \alpha +y\sin \alpha -p=0$ include an angle $\pi /4$ between them and meet the straight line $x\sin \alpha -y\cos \alpha =0$ in the same point, then the value of ${{a}^{2}}+{{b}^{2}}$is equal to A)            1      B)            2 C)            3      D)            4

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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