Prove the following

Question:

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

Solution:

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

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

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

( $\because$ equivalence class)

$x^{2}+y^{2}=2$

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