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AP EAMCET · Maths · Straight Lines

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

  1. A \(x^2+y^2-2 x y-2 x-2 y-1=0\)
  2. B ) \(x^2+y^2+2 x y+2 x+2 y+1=0\)
  3. C \(x^2+y^2+2 x y+2 x-2 y-1=0\)
  4. D \(x^2+y^2-2 x y+2 x-2 y+1=0\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(x^2+y^2+2 x y+2 x-2 y-1=0\)

Step-by-step Solution

Detailed explanation

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…