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Show that the relation `R`on 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 thepoint `Q`from the origin}, is an equivalence relation.Further show that the set of all points related to a point `P!=(0, 0)`is the circle passing through `P`with origin as centre. |
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Answer» `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.` `=>` 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) in R`. `:.` `R` is symmetric. Now, Let `(P, Q), (Q, S) in 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) in R`. `:. R` is transitive. As `R` is reflexive, summetric and transitive, `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 = r`, then the set of all points related to ` P` is at a distance of `r` from the origin. Hence, this set of points forms a circle with the centre as the origin and radius `r` and this circle passes through point `P`. |
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