Let R be a relation on the set A of points in a plane given byR= [ (P.Q): OP=OQ.O is the origin )(a) Show that R is an equivalence relation(b) Find [P], where P + (0,0)
Answers
Answer:
Step-by-step explanation:
R={(P,Q):distance of point P from the origin is the same as the distance of point Q from the origin}
Clearly, (P,P)∈R since the distance of point P from the origin is always the same as the distance of the same point P from the origin.
∴R is reflexive.
Now, let (P,Q)∈R.
⇒ the distance of point P from the origin is the same as the distance of point Q from the origin.
⇒ The distance of point Q from the origin is the same as the distance of point P from the origin.
⇒(Q,P)∈R
∴R is symmetric.
Now, let (P,Q),(Q,S)∈R.
⇒ The distance of points P and Q from the origin is the same and also, the distance of points Q and S from the origin is the same.
⇒ The distance of points P and S from the origin is the same.
⇒(P,S)∈R.
∴R is transitive.
Therefore, R is an equivalence relation.
The set of all points related to P
=(0,0) will be those points whose distance from the origin is the same as the distance of point P from the origin.
In other words, if O(0,0) is the origin and OP=k, then the set of all points related to P is at a distance of k from the origin.
Hence, this set of points forms a circle with the centre as the origin and this circle passes through point P.