Verified Solution • 2025–26 Session moderate

QJEEM21Ma15fe: JEE Main 2021 Mathematics - Class 11 JEEM21Ma15fe Solution: Let R = {(P, Q)

Let R = {(P, Q) | P and Q are at the same distance from the origin} be a relation, then the equivalence class of (1, −- 1) is the set :

Multiple Choice Options

  • A S={(x,y)∣x2+y2=2}S = \{ (x,y)|{x^2} + {y^2} = \sqrt 2 \}
  • B S={(x,y)∣x2+y2=2}S = \{ (x,y)|{x^2} + {y^2} = 2\}
  • C S={(x,y)∣x2+y2=1}S = \{ (x,y)|{x^2} + {y^2} = 1\}
  • D S={(x,y)∣x2+y2=4}S = \{ (x,y)|{x^2} + {y^2} = 4\}

Step-by-Step Verified Solution

Given R = {(P, Q) | P and Q are at the same distance from the origin}.

Then equivalence class of (1, −- 1) will contain al such points which lies on circumference of the circle of centre at origin and passing through point (1, −- 1).

i.e., radius of circle = 12+12=2\sqrt {{1^2} + {1^2}} = \sqrt 2

∴\therefore Required equivalence class of (S)

={(x,y)∣x2+y2=2}= \{ (x,y)|{x^2} + {y^2} = 2\} .

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