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AP EAMCET202022 Sep 2020Morning ShiftMathematicsStraight LinesActual

A point P(x, y) is such that the sum of squares of its distances from the co-ordinate axes is equal to the square of its distance from the line x-y=1 . Then the equation of the locus of P is

Options

  1. Ax^2+y^2-2 x y-2 x-2 y-1=0
  2. B) x^2+y^2+2 x y+2 x+2 y+1=0
  3. Cx^2+y^2+2 x y+2 x-2 y-1=0
  4. Dx^2+y^2-2 x y+2 x-2 y+1=0

Correct answer

C. x^2+y^2+2 x y+2 x-2 y-1=0

Step-by-step solution

It is given that the sum of squares of distance of point P(x, y) is equal to the square of its distance from the line x-y=1 , so aligned & x^2+y^2= (x-y-1)^2 2 2 x^2+2 y^2 & =x^2+y^2+1-2 x y-2 x+2 y & x^2+y^2+2 x y+2 x-2 y-1=0 aligned

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