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In the rectangular ...

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by bilbao » Mon May 31, 2010 11:57 am
Hello everyone,

I have not found the answer on the web, so I post my question.

In the rectangular coordinate system above, if OP < PQ, is the area of region OPQ greater than 48?

(1) The coordinates of P are (6,8).
(2) The coordinates of point Q are (13,0).

The answer is A.

My question is the following, I have the impression that answering A you do not consider the possibility that Q could be negative, isn't? In this case, we would have for instance Q=(-1,0) => PQ = sqrt(113) > OP BUT the area of OPQ would be (1*8)/2 = 4 => so less than 48...
If you consider the negative part of the axis as possible, the right answer would be C.

I am surely do not get something right. Could someone please help me?

Thanks a lot!!

Bil

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Source: — Data Sufficiency |
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by Haaress » Mon May 31, 2010 2:02 pm
The answer is A.

Lets have a perpendicular drop from P to the x axis and label the point of intersection as m. OM<MQ because OP <PQ.

If OP was equal to PQ then the area of traingle OPQ would have been 48 - double the area of triangle OPM = 2*1/2*6*8. But since PQ is greater . then the area will be greater given the two triangles share the same hieght, then the Area of triangle PMQ will larger than that of OPQ and thus OPQ is greater than 48.

Stmt 2 mentions nothing about the height, which is necessary in the calculation of the area.

Please let us know if you have any further questions on this. Thanks!
Last edited by Haaress on Mon May 31, 2010 2:26 pm, edited 1 time in total.
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by student22 » Mon May 31, 2010 2:12 pm
Guys, be careful when writing down the problem. This is a problem from the official guide, where it clearly shows you that Q is positive. Here's a screenshot.
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by kstv » Mon May 31, 2010 10:55 pm
Using bilbao's diagram
1)P(6,8) so OP = √6²+8² = 10
The perpendicular from P intersects the x axis at D, PD = 8 and DO = 6
PQ > 10 so DQ > √10²- 8² or > 6
OQ > 12
so area of POQ > (8*12)/2 > 48
Sufficient.
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