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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2004

  • question_answer
    The locus of the point\[(x,\text{ }y)\]which is equidistant from the points\[(a+b,b-a)\]and \[(a-b,a+b)\]is:

    A)  \[ax=by\]         

    B) \[ax+by=0\]

    C)  \[bx+ay=0\]     

    D)        \[bx-ay=0\]

    E)  \[\frac{x}{a}+\frac{y}{b}=1\]

    Correct Answer: D

    Solution :

    Let\[P(x,\text{ }y)\]is equidistant from the points \[A(a+b,b-a)\]and\[(a-b,a+b)\] \[\therefore \]  \[P{{A}^{2}}=P{{B}^{2}}\] \[\Rightarrow \]               \[{{(a+b-x)}^{2}}+{{(b-a-y)}^{2}}\] \[={{(a-b-x)}^{2}}+{{(a+b-y)}^{2}}\] \[\Rightarrow \]               \[bx-ay=0\]


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