Question
Express $\left\{(x, y) / x^2+y^2=100\right.$, where $\left.x, y \in W\right\}$ as a set of ordered pairs.

Answer

$
\left\{(x, y) / x^2+y^2=100, \text { where } x, y \in W\right\}
$
We have, $x^2+y^2=100$
When $x=0$ and $y=10$,
$
x^2+y^2=0^2+10^2=100
$
When $x=6$ andy $=8$,
$
x^2+y^2=6^2+8^2=100
$
When $x=8$ and $y=6$,
$
x^2+y^2=8^2+6^2=100
$
When $x =10$ and $y =0$,
$
x^2+y^2=10^2+0^2=100
$
$\therefore$ Set of ordered pairs $=\{(0,10),(6,8),(8,6),(10,0)\}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free