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In the figure above

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vrajesh.dave
 Post subject: In the figure above
 Post Posted: Sun Aug 23, 2009 9:55 pm 
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Course Students


Posts: 31
Hi -

I tried to search the forum for this question, but could not find it [maybe I missed it].
Image
OA: B.

I got the part where I find the radius of the semi-circle.

Using
distance = sqrt((y2 - y1)^2 + (x2 - x1)^2)

So I know radius is 2. But how do we find the coordinates of s,t
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srinu84us
 Post subject: Re: In the figure above
 Post Posted: Mon Aug 24, 2009 1:54 am 
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Students


Posts: 1
distance from (0,0) to (s,t) should be equal to (0,0) to (-sqrt3, 1) hence s^2 + t^2 = 4
since it is a right angle, m1m2 = -1; m1 = t/s ; m2 = -1/sqrt(3) substituting we will get s = 1
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vrajesh.dave
 Post subject: Re: In the figure above
 Post Posted: Mon Aug 24, 2009 7:35 am 
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Course Students


Posts: 31
I am assuming that you mean sqrt((0-s)^2 + (0 - t)^2) = 4

what do m1 and m2 represent??

And I still do not understand how you got t/s = 1.

Can you please explain?
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nimish.tiwari
 Post subject: Re: In the figure above
 Post Posted: Tue Aug 25, 2009 2:58 am 
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Students


Posts: 13
vrajesh.dave wrote:
I am assuming that you mean sqrt((0-s)^2 + (0 - t)^2) = 4

what do m1 and m2 represent??

And I still do not understand how you got t/s = 1.

Can you please explain?


putting values of m1 & m2 as mentioned by srinu84us into m1*m2=-1, you'll get t = s * (sqrt(3))
Put this back into t^2 + s^2 = 4 and solve for s. You'll get s = +1 or -1.
since point (s,t) is in 1st quadrant so s has to be positive. Hence, s=1. Alternately, you can also conclude that s=1 since -1 is not an option enlisted.
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RonPurewal
 Post subject: Re: In the figure above
 Post Posted: Fri Sep 25, 2009 10:57 pm 
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ManhattanGMAT Staff


Posts: 7146
here's a huge thread on this problem

gmat-prep-geometry-2-t2493.html
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