If - - x-y - x 2 - y 2 -1-x-y- R - Then- - is

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Standard XII

Mathematics

Transpose of a Matrix

If ρ = x,y ...

Question

If $ρ = {(x, y) | x^{2} + y^{2} = 1; x, y \in R}$ . Then, $ρ$ is

Reflexive

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Symmetric

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Transitive

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Anti-symmetric

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Solution

The correct option is D Symmetric
Obviously, the relation is not reflexive and transitive but it is symmetric, because
$x^{2} + x^{2} = 2 x^{2} \neq 1$
and $x^{2} + y^{2} = 1, y^{2} + z^{2} = 1 \neq> x^{2} + z^{2} = 1$
$\Rightarrow x^{2} + y^{2} = 1$
$\Rightarrow y^{2} + x^{2} = 1$

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If ρ={(x,y)|x2+y2=1;x,y∈R}. Then, ρ is

The correct option is D SymmetricObviously, the relation is not reflexive and transitive but it is symmetric, becausex2+x2=2x2≠1and x2+y2=1,y2+z2=1≠>x2+z2=1⇒x2+y2=1⇒y2+x2=1

If $ρ = {(x, y) | x^{2} + y^{2} = 1; x, y \in R}$ . Then, $ρ$ is

The correct option is D Symmetric
Obviously, the relation is not reflexive and transitive but it is symmetric, because
$x^{2} + x^{2} = 2 x^{2} \neq 1$
and $x^{2} + y^{2} = 1, y^{2} + z^{2} = 1 \neq> x^{2} + z^{2} = 1$
$\Rightarrow x^{2} + y^{2} = 1$
$\Rightarrow y^{2} + x^{2} = 1$