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JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2011

  • question_answer
        The   point   P   is   equidistant   from A (1, 3), B \[(-3,5)\]and\[C(5,-1)\]. Then, PA is equal to

    A)  \[5\sqrt{5}\]                     

    B)  \[5\]

    C)  \[5\sqrt{10}\]                  

    D)  \[25\]

    Correct Answer: C

    Solution :

                    Given  points  are \[A(1,3),B(-3,5)\]and\[C(5,-1)\]Mid-piont of AB is\[(-1,4),\]slope of\[AB=-\frac{1}{2}\]. Perpendicular bisector of AB,\[y-4=2(x+1)\] \[\Rightarrow \]               \[2x-y+6=0\]                   ...(i) Similarly, perpendicular bisector of AC is \[x-y-2=0\]                       ...(ii) Point of intersection of Eqs. (i) and (ii) is \[P=(-8,-10)\] Then, \[PA=\sqrt{{{(1+8)}^{2}}+{{(3+10)}^{2}}}\] \[=\sqrt{81+169}=5\sqrt{10}\]


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