GMAT Prep Triangle and Coordinate Plane

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GMAT Prep Triangle and Coordinate Plane

by tonebeeze » Sun May 22, 2011 11:58 am
Can someone walk me through this problem and help me visualize it. Thanks.


For any triangle T in the xy-xoordinate 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)

OA = B
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by Ashley@VeritasPrep » Sun May 22, 2011 12:21 pm
Hi there,

The idea on this problem is that the point being defined as the "center" of the triangle is the point whose x-coordinate is the average of the three vertices' x-coordinates, and whose y-coordinate is the average of the three vertices' y-coordinates. I've put the word "center" in quotes because this point will not necessarily look (if you actually plot all the points) like it's definitively in the center of the triangle. The point being described actually has a name -- the centroid -- and in the physical world, if the triangle were a very thin plate of material, say -- it's the point in the interior of the triangle where you'd have to situate your pencil tip (sticking straight up from underneath) if you wanted this triangle to balance perfectly on your pencil tip. Put the pencil tip anywhere else underneath the triangle, and the triangle will not balance. You don't actually need to know any of this to solve the problem, but it makes it more interesting to think about :)

In this problem, we've already been told that the center is (3,2), so 3 must be the average of our vertices' x-coordinates, and 2 must be the average of our vertices' y-coordinates. We know the coordinates of two of our vertices -- (0,0) and (6,0) -- and let's call the other one (a,b). Then 3 must be the average of 0, 6, and a, so

(0+6+a)/3 = 3 --> 6+a = 9 --> a = 3

and 2 must be the average of 0, 0, and b, so

(0+0+b)/3 = 2 --> b = 6

So the coordinates of the third vertex are (3,6).

Hope that helps!

Ashley Newman-Owens
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by tonebeeze » Sun May 22, 2011 12:42 pm
Very clear and simple explanation. Thank you!
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