JEE MAIN - Mathematics (2021 - 26th February Morning Shift - No. 6)
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 :
$$S = \{ (x,y)|{x^2} + {y^2} = \sqrt 2 \} $$
$$S = \{ (x,y)|{x^2} + {y^2} = 2\} $$
$$S = \{ (x,y)|{x^2} + {y^2} = 1\} $$
$$S = \{ (x,y)|{x^2} + {y^2} = 4\} $$
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)
$$ = \{ (x,y)|{x^2} + {y^2} = 2\} $$.
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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