Prove that $pTq \longleftrightarrow |p| = |q|$ is Equivalence Relation on $A$ set of all point in the plane I want to prove that this relation is equivalence relation on A


*

*$A$ set of all points in the plane

*$pTq \longleftrightarrow |p| = |q|$ , |p| is the distance from origin.


about transitivity,  there are counter-examples?
for reflexivity is obvious, $(x,x)$ the distance will be the same.
for symmetry $(x,y)\in R , (y,x) \in R$ the distances are the same.
if its Equivalence Relation what are the equivalence classes? and partition set?

I would like to get some suggestions.

Reflexivity:
$$\forall\,(a,b)\in\Bbb R^2\;,\;\;\sqrt{a^2+b^2}=\sqrt{a^2+b^2}\implies (a,b)T(a,b)$$
Symmetry:
$$(a,b)T(c,d)\implies \sqrt{a^2+b^2}=\sqrt{c^2+d^2}\implies\sqrt{c^2+d^2}=\sqrt{a^2+b^2}\implies (c,d)T(a,b) $$
Transitivity:
$$(a,b)T(c,d)\;\wedge\;(c,d)T(x,y)\implies \sqrt{a^2+b^2}=\sqrt{c^2+d^2}=\sqrt{x^2+y^2}\implies$$
$$(a,b)T(x,y)$$
The equivalence classes here : $\;x^2+y^2=R^2 ...$
