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GMAT- PREP x-coordinate of the point on circle
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crazykrans
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Didn't find this in my search in this forum. I need the method or a pointer to a previous post . Thanks.
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muzali
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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.
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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crazykrans
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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.
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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crazykrans
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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 ?)
Is there a better method ?)
