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JEE Main & Advanced Mathematics Straight Line Question Bank Self Evaluation Test - Straight Lines & Pair Striaght Lines

  • question_answer
    If the point \[P(x,y)\] is equidistant from points\[A(a+b,b-a)\] and \[B(a-b,a+b)\], then

    A) \[ax=by\]

    B) \[bx=ay\] and P ca be (a, b)

    C) \[{{x}^{2}}-{{y}^{2}}=2(ax+by)\]

    D) None of the above

    Correct Answer: B

    Solution :

    [b] We, have, \[PA=PB\Rightarrow {{(PA)}^{2}}={{(PB)}^{2}}\] \[\Rightarrow {{[x-(a+b)]}^{2}}+{{[y-(b-a)]}^{2}}\] \[={{[x-(a-b)]}^{2}}+{{[y-(a+b)]}^{2}}\] \[\Rightarrow {{[(x-a)-b]}^{2}}+{{[(y-b)+a]}^{2}}\] \[={{[(x-a)+b]}^{2}}+{{[(y-b)-a]}^{2}}\] \[\Rightarrow {{[(x-a)+b]}^{2}}-{{[(x-a)-b]}^{2}}\] \[={{[(y-b)+a]}^{2}}-{{[(y-b)-a]}^{2}}\] \[\Rightarrow 4b(x-a)=4a(y-b)\Rightarrow bx=ay\]  ? (i) Also, P (a, b) satisfies the condition (i), so that P can be (a, b).


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