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  • Show that the relation R in the set A of points in a plane given by R = {(P, Q) : distance of the point P from the origin is same as the distance of the point Q from the origin}, is an equivalence relation. Further, show that the set of all points related

    Q.11 Show that the relation R in the set A of points in a plane given by
R = \{(P, Q) : \;distance \;of \;the\; point\; P\; from \;the \;origin \;is \;same \;as \;the\; distance \;of \;the \;point \;Q \;from \;the \;origin\}, is an equivalence relation. Further, show that the set of
all points related to a point P \neq (0, 0) is the circle passing through P with origin as
centre.

Answers (2)

best_answer

R = \{(P, Q) : \;distance \;of \;the\; point\; P\; from \;the \;origin \;is \;same \;as \;the\; distance \;of \;the \;point \;Q \;from \;the \;origin\}

The distance of point P from the origin is always the same as the distance of same point P from origin i.e.(P,P)\in R

 \therefore R is reflexive.

Let (P,Q)\in R i.e. the distance of the point P from the origin is the same as the distance of the point Q from the origin.

this is the same as distance of the point Q from the origin is the same as the distance of the point P from the origin i.e. (Q,P)\in R

\thereforeR is symmetric.

Let    (P,Q)\in R     and    (Q,S)\in R 

       i.e. distance of the point P from the origin is same as the distance of the point Q from the origin, and also the distance of the point Q from the origin is same as the distance of the point S from the origin.

We can say that distance of point P, Q, S  from the origin is the same. Means distance of point P from the origin is the same as the distance of point S from origin i.e.  (P,S)\in R

\therefore  R is transitive.

Hence, R  is an equivalence relation.

 

The set of all points related to a point P \neq (0, 0) are points whose distance from the origin is the same as the distance of point P from the origin.

In other words, we can say there be a point O(0,0) as origin and distance between point O and point P be k=OP then set of all points related to P is at distance k from the origin.

Hence, these set of points form circle with the centre as the origin and this circle passes through point P.

 

 

Posted by

seema garhwal

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R = \{(P, Q) : \;distance \;of \;the\; point\; P\; from \;the \;origin \;is \;same \;as \;the\; distance \;of \;the \;point \;Q \;from \;the \;origin\}

The distance of point P from origin is always same as distance of same point P from origin i.e.(P,P)\in R

 \therefore R is reflexive.

Let (P,Q)\in R i.e. distance of the point P from the origin is same as the distance of the point Q from the origin.

this is same as : distance of the point Q from the origin is same as the distance of the point P from the origin i.e. (Q,R)\in R

\thereforeR is symmetric.

Let    (P,Q)\in R     and    (Q,S)\in R 

       i.e. distance of the point P from the origin is same as the distance of the point Q from the origin, and aslo  distance of the point Q from the origin is same as the distance of the point S from the origin.

We can say that distance of point P,Q,S  from origin is same.Means distance of point P from origin is same as distance of point S from origin i.e.  (P,S)\in R

\therefore  R is transitive.

Hence, R  is an equivalence relation.

 

The set of all points related to a point P \neq (0, 0) are points whose distance from origin is same as distance of point P from origin.

In other words we can say there be a point O(0,0) as origin and  distance between point O and point P be k=OP then set of all points related to P is at distance k from origin.

Hence, these set of points form circle with centre as origin and this circle passes through point P.

 

 

Posted by

seema garhwal

View full answer

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