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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 :

JEE · Math · previous-year question

  1. A.$$S = \{ (x,y)|{x^2} + {y^2} = \sqrt 2 \} $$
  2. B.$$S = \{ (x,y)|{x^2} + {y^2} = 2\} $$correct
  3. C.$$S = \{ (x,y)|{x^2} + {y^2} = 1\} $$
  4. D.$$S = \{ (x,y)|{x^2} + {y^2} = 4\} $$

Answer

B. $$S = \{ (x,y)|{x^2} + {y^2} = 2\} $$

Explanation

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 = $$\sqrt {{1^2} + {1^2}} = \sqrt 2 $$ $$ \therefore $$ Required equivalence class of (S) $$ = \{ (x,y)|{x^2} + {y^2} = 2\} $$.

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