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# In the figure above, points P and Q

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Manager
Joined: 05 Aug 2008
Posts: 89
Schools: McCombs Class of 2012
In the figure above, points P and Q [#permalink]

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24 Dec 2008, 16:45
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SVP
Joined: 07 Nov 2007
Posts: 1761
Location: New York
Re: Points in a circle [#permalink]

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25 Dec 2008, 00:37
smarinov wrote:
Attachment:
Q1.JPG

slope of the line OP = 1-0/-SQRT(3)-0 = - 1/SQRT(3)

slope of the line OQ, which is perpendicular to OP = sqrt(3) = t-0/s-0 = t/s

$${(-sqrt(3))}^{2} + 1^2=4 =OQ$$
OQ= 4=
$${s*sqrt(3)}^2+ s^2$$
--> s=1
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Re: Points in a circle   [#permalink] 25 Dec 2008, 00:37
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# In the figure above, points P and Q

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