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Triangle T

by zagcollins » Mon Jul 28, 2008 9:03 am
For any triangle T in the xy-coordinate plane, the center of T is defined to be the point whose x-coordinate is the average (arithmetic mean) of the x-coordinates of the vertices of T and whose y-coordinate is the average of the y-coordinates of the vertices of T.If a certain triangle has vertices at the points (0,0) and (6,0) and center at the point (3,2), what are the coordinates of the remaining vertex?

A.(3,4)
B.(3,6)
C.(4,9)
D.(6,4)
E.(9,6)

The OA is B...the source is GPREP! :lol:
i believe malolakrupa is spot on with the explanation!
Last edited by zagcollins on Tue Jul 29, 2008 2:49 am, edited 2 times in total.
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by ricky » Mon Jul 28, 2008 9:25 am
Is the ans B?
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by ramyaravindran » Mon Jul 28, 2008 9:30 am
The center of the triangle must be equidistant from the vertices. The only option that fits this scenario is D.
Distance between (0,0) and (3,2) is SQRT(13)
Distance between (6,0) and (3,2) is SQRT(13)
Distance between (6,4) and (3,2) is also SQRT(13)
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by malolakrupa » Mon Jul 28, 2008 9:53 am
For X - coordinate ,

(0+6+X) / 3 = 3 , which gives X = 3

For Y coordinate
(0 + 0 + Y) / 3 = 2 which gives Y = 6

Hence B is the answer
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by parallel_chase » Mon Jul 28, 2008 11:33 am
I think this question is incorrect or you are missing some piece of information.
Since in GMAT one cannot have 2 correct answers, I think this question is wrong, because:
if you go by the arithmetic mean the answer is (3,6) but if you go by the fundamentals of triangle i.e center of the triangle is equidistant from all its vertices then the answer is (6,4). But then the averages or arithmetic mean becomes redundant.

Whats the OA? and above all the source.

Thanks
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