Let P-x-y-x2-y2-1-x-y- R- then P is

Explanation for the correct option.Finding the relation of PGiven: x^2 + y^2 = 1If (x,x)∈ R then R is ReflexiveIf (x,y)∈ R then (y,x) ∈ Rthen R is SymmetricIf ...

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Byju's Answer

Standard IX

Mathematics

Symmetric Relations

Let P=x,y|x2+...

Question

Let $P = \{(x, y) | x^{2} + y^{2} = 1; x, y \in R\}$ . then $P$ is

Reflexive

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Symmetric

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Transitive

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Antisymmetric

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Solution

The correct option is B
Symmetric

Explanation for the correct option.
Finding the relation of $P$
Given: $x^{2} + y^{2} = 1$
If $(x, x) \in R then R$ is Reflexive
If $(x, y) \in R then (y, x) \in R$ then $R$ is Symmetric
If $(x, y) \in R and (y, z) \in R then (x, z) \in R$ then $R$ transitive.
From the above definition it is clear that $R$ is Symmetric.
Since $x^{2} + y^{2} = 1$
$\Rightarrow y^{2} + x^{2} = 1$
Hence option (B) is the correct answer.

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Similar questions

Q. Let

P = {(x, y) | x^{2} + y^{2} = 1, x, y \in R}

. Then

P

Let P = {(x,y) $x^{2} + y^{2} = 1, x, y \in R}$ . Then P is.

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 :

Q. Let

P = {(x, y) | x^{2} + y^{2} = 1, x, y \in R}

. Then

P

Q. Let

A = {X = {(x, y, z)}^{T} : P X = O and x^{2} + y^{2} + z^{2} = 1},

where

P = ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & 1 - 2 & 3 & - 4 1 & 9 & - 1 \end{matrix} ⎤ ⎥ ⎦

, then the set

A

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Let P=(x,y)|x2+y2=1;x,y∈R. then P is

Let $P = \{(x, y) | x^{2} + y^{2} = 1; x, y \in R\}$ . then $P$ is