Question

Please help me

Show that the point (7/25,24/25) is on the unit circle

Answer 1

So just a^2+b^2=c^2?

Answer 2

yea

Answer 3

Oh wait, I think you got the wrong problem. This is the point you have to find on the unit circle \[(\frac{ 7 }{ 25 },\frac{ 24 }{ 25 })\]

Answer 4

@Janu16

Answer 5

no i got it right. but heres the other way

The center of the unit circle is (0,0). If the distance between the center and the given point turns out to be 1, then the point is on the unit circle.
The distance of the given point from (0,0) in this case works out to be
\[\sqrt{\left(\begin{matrix}7 \\ 25\end{matrix}\right)^2+\left(\begin{matrix}24 \\ 25\end{matrix}\right)^2}\]
\[=\sqrt{49+576}\div625\]
=1

Answer 6

@ohohaye

Answer 7

oh, I'm sorry, I've never seen someone solve it how you were solving it in the first part

Answer 8

but didyou get 2nd part?

Answer 9

@ohohaye

Answer 10

I'm confused as to this part

Answer 11

@Janu16

Answer 12

so 49+576 is 625 and when you divide 625 by 625 its 1

Answer 13

@ohohaye

Answer 14

Ok, thank you