Question

Find all of the points of the form (x, −x) which are 3 units from the origin.

Answer 1

with smaller x-value and larger x-value

Answer 2

(x , -x)

This limits us to the line y = -x. Do you see this?

Answer 3

no there has to be 2 answers..

Answer 4

You did not answer my question. Please answer it. Don't be jumping ahead to what you think the answers are. Let's prove it!

Answer 5

No I do not see how you got your answer, please show me then

Answer 6

Points look like this (x,y)
If we restrict ALL points to a subset that looks like this (x,-x), we have defined the line y = -x

Holding them next to each other, what changed?

(x,y)
(x,-x)

Answer 7

y changes to -x

Answer 8

Thus, y = -x. NOW do you see where that came from?

Answer 9

yes but that still doesn't answer the question because I need a (x,y) = a smaller x value and an (x,y) = a larger x value

Answer 10

You are being very jumpy. We need to understand things. Stop looking at the answers. Let's think about what we are doing.

We know our answer(s) lie in the line y = -x. You have very astutely observed that this is infinitely many points. That just won't do! Let's go check the problem statement and see if there is another hint.

Aha! "3 units from the origin"

This means our solution(s) must lie on the circle \(x^{2} + y^{2} = 3^{2}\). Do you see this and agree with it?

Answer 11

yes so i would put x^2+(-x)^2=3^2 right?

Answer 12

I will allow that little jump. You jumped to the next step and did not attempt to jumpt to the end. Good work!

Yes. Now, solve for x.

Answer 13

\[x=+/-\sqrt{18}/4\]

Answer 14

What an odd-looking result.

x^2 + x^2 = 2x^2 = 3^2

x^2 = (3^2)/2

\(x = \pm\dfrac{3}{\sqrt{2}}\)

Why would you expand 3^3 to 9, if you know you are about to find its square root and end up back at 3?

Okay, these are the x-values. What are the y-values associated with them? This should be very easy.

Hint: (x,-x)

Answer 15

i have no idea... im still alittle lost on how you got the square root on the bottom for 2

Answer 16

Same way you did. How did you get 18 in the numerator and 4 in the denominator? I'm guessing you made an effort to rationalize the denominator. That's fine, but you also missed that \(\sqrt{3^{2}} = 3\). You still have \(\sqrt{9}\)

Answer 17

okay i see it now. so y values are just -x then?

Answer 18

That's it. Remind yourself how the solutions are on both

y = -x

and

x^2 + y^2 = 3^2

Answer 19

okay my last question then would be if its asking for a smaller x-value and a larger x-value would I put... (x,y)= \[-3/\sqrt{2}, -x\] for smaller x-value and (x,y)= \[3/\sqrt{2} , -x)\] for larger x-value?

Answer 20

I suppose. That is a rather odd request. Also, fill in the value for -x. It's not -x anymore. We know its value!

Answer 21

what? you just said that the y value is -x.....

Answer 22

@spectrum I just need the y value and then im good

Answer 23

tk is helping

Answer 24

well I would love an answer now....

Answer 25

You already have the y-value. Why do you still seek it.

If x = sqrt(3)/2 and y = -x, then y = -sqrt(3)/2.

Answer 26

ohhh... got it, thank you!