Prove the following


Let $\mathrm{R}=\{\mathrm{P}, \mathrm{Q}) \mid \mathrm{P}$ and $\mathrm{Q}$ are at the same distance from the origin $\}$ be a relation, then the equivalence class of $(1,-1)$ is the set:

  1. (1) $S=\left\{(x, y) \mid x^{2}+y^{2}=1\right\}$

  2. (2) $S=\left\{(x, y) \mid x^{2}+y^{2}=4\right\}$

  3. (3) $S=\left\{(x, y) \mid x^{2}+y^{2}=\sqrt{2}\right\}$

  4. (4) $S=\left\{(x, y) \mid x^{2}+y^{2}=2\right\}$

Correct Option: , 4


$\mathrm{P}(\mathrm{a}, \mathrm{b}), \mathrm{Q}(\mathrm{c}, \mathrm{d}), \mathrm{PO}=\mathrm{QO}$

$\Rightarrow a^{2}+b^{2}=c^{2}+d^{2}$

$\mathrm{R}(\mathrm{x}, \mathrm{y}) \quad \mathrm{s}=(1,-1) \Rightarrow \mathrm{RO}=\mathrm{SO}$

( $\because$ equivalence class)


Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now