GMAT Prep - Tough PS Triangle, Coordinate Geometry - The Beat The GMAT Forum

GMAT Prep - Tough PS Triangle, Coordinate Geometry

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albertrahul: Senior | Next Rank: 100 Posts; Posts: 83; Joined: Wed Jun 04, 2008 3:30 am; Thanked: 4 times

by albertrahul » Wed Jul 02, 2008 8:26 pm

Preety good question, though I got it wrong in first attempt.
A is not OA. I'll post OA after discussion.

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Source: Beat The GMAT — Quantitative Reasoning | Wed Jul 02, 2008 8:26 pm

simpdimp: Newbie | Next Rank: 10 Posts; Posts: 7; Joined: Sat Jan 19, 2008 7:26 pm; Thanked: 1 times

by simpdimp » Wed Jul 02, 2008 10:33 pm

Looks like a pretty simple question, let me know if I got it right!

(6+0+x)/3 = 3 and (0+0+y)/3 = 2 --> (3,6)

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asigheartau: Junior | Next Rank: 30 Posts; Posts: 18; Joined: Sun Jun 22, 2008 11:25 am; Location: NY; Thanked: 1 times

by asigheartau » Fri Jul 04, 2008 9:24 am

simpdimp wrote:Looks like a pretty simple question, let me know if I got it right!

(6+0+x)/3 = 3 and (0+0+y)/3 = 2 --> (3,6)

If the values of x are the result of the average of the x vertices, shouldn't it be (6+0)/2=3 in which case y would be (0+2+Y)/3=??

Can you please detail your solution?

Alin Sigheartau

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albertrahul: Senior | Next Rank: 100 Posts; Posts: 83; Joined: Wed Jun 04, 2008 3:30 am; Thanked: 4 times

by albertrahul » Sat Jul 05, 2008 6:50 am

Answer is (3,6).
This is actually a pretty simple question, but verbiage is confusing.

We've been told that values of x and y are average of respective vertices so value for x coordinate will be (6+0+x)/3 = 3.

Hope that helps!

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tolmar: Senior | Next Rank: 100 Posts; Posts: 39; Joined: Sat Jul 12, 2008 5:18 am

by tolmar » Sun Jul 13, 2008 12:00 pm

(3,6) to my opinion. This question is just long and cumbersome.

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