JEE Main & Advanced Mathematics Straight Line Question Bank Critical Thinking

  • 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

    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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