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} = 1}

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S = {(x, y) | x^{2} + y^{2} = 4}

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S = {(x, y) | x^{2} + y^{2} = \sqrt{2}}

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S = {(x, y) | x^{2} + y^{2} = 2}

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Solution

The correct option is D $S = {(x, y) | x^{2} + y^{2} = 2}$
Let $P (a, b)$ and $Q (c, d)$ be two points.
$O P = O Q$
$\Rightarrow a^{2} + b^{2} = c^{2} + d^{2}$
$R (x, y), S (1, - 1)$
$\Rightarrow O R = O S (∵$ equivalence class $)$
$\Rightarrow x^{2} + y^{2} = 2$

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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 :

The correct option is D S={(x,y) | x2+y2=2}Let P(a,b) and Q(c,d) be two points. OP=OQ ⇒a2+b2=c2+d2 R(x,y), S(1,−1) ⇒OR=OS (∵ equivalence class ) ⇒x2+y2=2

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 :

The correct option is D $S = {(x, y) | x^{2} + y^{2} = 2}$
Let $P (a, b)$ and $Q (c, d)$ be two points.
$O P = O Q$
$\Rightarrow a^{2} + b^{2} = c^{2} + d^{2}$
$R (x, y), S (1, - 1)$
$\Rightarrow O R = O S (∵$ equivalence class $)$
$\Rightarrow x^{2} + y^{2} = 2$