10th Class Mathematics Solved Paper - Mathematics-2016 Outside Delhi Set-I

  • question_answer 13) If the point \[P(x,y)\] is equidistant from the points \[A(a+b,b-a)\] and \[B(a-b,a+b)\]. Prove that \[bx=ay\].

    Answer:

    Since, P us equidistant from points A and B,
    \[\therefore \]                  \[PA=PB\]
    Or,                    \[{{(PA)}^{2}}={{(PB)}^{2}}\]
    \[{{(a+b-x)}^{2}}+{{(b-a-y)}^{2}}={{(a-b-x)}^{2}}+{{(a+b-y)}^{2}}\]
    \[{{(a+b)}^{2}}+{{x}^{2}}-2ax-2bx+{{(b-a)}^{2}}+{{y}^{2}}-2by+2ay\]\[={{(a-b)}^{2}}+{{x}^{2}}-2ax+2bx+{{(a+b)}^{2}}+{{y}^{2}}-2ay-2by\]
                   \[-2bx+2ay=2bx-2ay\]
                            \[4ay=4bx\]
                            \[ay=bx\]
                            \[bx=ay\]                                  Hence Proved.


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