Let R = {(P, Q) | P and Q are at the same distance from the origin} be a relation, then the equivalence class of (1, $-$1) is the set :
Solution
Given R = {(P, Q) | P and Q are at the same distance from the origin}.<br><br>Then equivalence class of (1, $-$1) will contain al such points which lies on circumference of the circle of centre at origin and passing through point (1, $-$1).<br><br>i.e., radius of circle = $\sqrt {{1^2} + {1^2}} = \sqrt 2$<br><br>$\therefore$ Required equivalence class of (S)<br><br>$= \{ (x,y)|{x^2} + {y^2} = 2\}$.
About this question
Subject: Mathematics · Chapter: Sets, Relations and Functions · Topic: Sets and Operations
This question is part of PrepWiser's free JEE Main question bank. 195 more solved questions on Sets, Relations and Functions are available — start with the harder ones if your accuracy is >70%.