Show that the relation R in the set A of points in a plane given by
Question:

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 point related to a point $P \neq(0,0)$ is the circle passing through $P$ with origin as centre.

Solution:

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

Clearly, $(P, P) \in R$ 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 $(P, Q) \in R$.

$\Rightarrow$ The distance of point $P$ from the origin is the same as the distance of point $Q$ from the origin.

$\Rightarrow$ The distance of point $Q$ from the origin is the same as the distance of point $P$ from the origin.

$\Rightarrow(Q, P) \in R$

∴R is symmetric.

Now,

Let $(P, Q),(Q, S) \in R$.

$\Rightarrow$ 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.

$\Rightarrow$ The distance of points $\mathrm{P}$ and $\mathrm{S}$ from the origin is the same.

$\Rightarrow(P, S) \in R$

∴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.

 

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