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New post 12 Aug 2008, 10:32
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Re: DS- GMATPREP - coordinate geometry [#permalink]

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New post 12 Aug 2008, 10:52
s=1, right? I'll explain it if I'm right.

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New post 12 Aug 2008, 10:55
yes, you are correct. Please explain

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New post 12 Aug 2008, 11:04
mba9now wrote:
yes, you are correct. Please explain



If point P has coordinates (-root3,1), then the hypotonese, which is also the radius comes out to be 2. The tricky part of this question is that you are led to believe that the radius is root3, which is not correct. The second trick of this question is that it leads you to believe that the y-axis bisects the 90 degree angle at exactly 45 degrees, which is also not correct.

So, to generate a radius of 2 at point Q, the coordinates of P have to be flipped. I can draw it, but I don't have a lot of time now. I hope this helps. Drawing it certainly clarifies it.

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New post 12 Aug 2008, 11:10
but (+root3 , 1) will also get you a radius of 2.

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New post 12 Aug 2008, 11:24
mba9now wrote:
but (+root3 , 1) will also get you a radius of 2.



Yes exactly. We want it to give a radius of 2 because it lies on the circle.

OP = OQ because both point are on the circle.

On the left triangle, we have a rise of 1 and a run of root3. For POQ to be 90 degrees, we will need the rise and run of the triangle to the right be the opposite of the rise and run of the triangle to the left. Thus, the coordinates of point Q are (1,root3).

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Re: DS- GMATPREP - coordinate geometry [#permalink]

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New post 12 Aug 2008, 12:14
mba9now wrote:
Please explain your answer. OA to follow...

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radius=2
becaue OP and OQ perpendicular -- (s,t) must be in the ratio 1:sqrt(3)
right angle triangle. radius r=2
s=1 t=sqrt(3)
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Re: DS- GMATPREP - coordinate geometry   [#permalink] 12 Aug 2008, 12:14
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