anonymous
  • anonymous
Find all of the points of the form (x, −x) which are 3 units from the origin.
Algebra
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
with smaller x-value and larger x-value
tkhunny
  • tkhunny
(x , -x) This limits us to the line y = -x. Do you see this?
anonymous
  • anonymous
no there has to be 2 answers..

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More answers

tkhunny
  • tkhunny
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!
anonymous
  • anonymous
No I do not see how you got your answer, please show me then
tkhunny
  • tkhunny
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)
anonymous
  • anonymous
y changes to -x
tkhunny
  • tkhunny
Thus, y = -x. NOW do you see where that came from?
anonymous
  • anonymous
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
tkhunny
  • tkhunny
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?
anonymous
  • anonymous
yes so i would put x^2+(-x)^2=3^2 right?
tkhunny
  • tkhunny
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.
anonymous
  • anonymous
\[x=+/-\sqrt{18}/4\]
tkhunny
  • tkhunny
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)
anonymous
  • anonymous
i have no idea... im still alittle lost on how you got the square root on the bottom for 2
tkhunny
  • tkhunny
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 \(\sqrt{9}\)
anonymous
  • anonymous
okay i see it now. so y values are just -x then?
tkhunny
  • tkhunny
That's it. Remind yourself how the solutions are on both y = -x and x^2 + y^2 = 3^2
anonymous
  • anonymous
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?
tkhunny
  • tkhunny
I suppose. That is a rather odd request. Also, fill in the value for -x. It's not -x anymore. We know its value!
anonymous
  • anonymous
what? you just said that the y value is -x.....
anonymous
  • anonymous
@spectrum I just need the y value and then im good
anonymous
  • anonymous
tk is helping
anonymous
  • anonymous
well I would love an answer now....
tkhunny
  • tkhunny
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.
anonymous
  • anonymous
ohhh... got it, thank you!

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