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GMAT- PREP x-coordinate of the point on circle

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Source: — Problem Solving |
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by muzali » Fri Dec 26, 2008 9:22 am
Several ways of solving:

1. Lines OP and OQ are perpendicular, so product of their slopes will be -1.
slope of OP*slope of OP=-1

slope of OP = -(1/root3)
=>Slope of OQ = root3 = t/s. So s = 1

2.Can also use the 30-60-90 traingles method to see that s is 1.
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by crazykrans » Fri Dec 26, 2008 9:52 am
Muzali - I'm sorry, may be I'm missing something here.
If slope of OQ = sq root 3 = t/s , how is s=1 and not t/sq root 3 ?
Q could be either (sq root 3, 3) or (1, sq root 3.) or some other point.

Appreciate if anyone can throw some light on this.
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by crazykrans » Fri Dec 26, 2008 9:55 am
Ok, I got it select the point Q such that the radius is same as OP which is 2. So Q(1, sq root 3) fits it nicely.
Is there a better method ?)
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