AP EAMCET Engineering 22 Sep 2020 Shift 1 Paper

Section: Mathematics

Question : 19 of 160

Marks: +1, -0

A point P(−3,−2) 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

x2+y2−2xy−2x−2y−1 = 0
x2+y2+2xy+2x+2y+1 = 0
x2+y2+2xy+2x−2y−1 = 0
x2+y2−2xy+2x−2y+1 = 0

Validate

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