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# On the coordinate plane (6, 2) and (0, 6) are the endpoints

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Manager
Joined: 29 Nov 2011
Posts: 80
On the coordinate plane (6, 2) and (0, 6) are the endpoints [#permalink]

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24 Jul 2012, 08:03
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Difficulty:

65% (hard)

Question Stats:

54% (04:21) correct 46% (02:04) wrong based on 39 sessions

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On the coordinate plane (6, 2) and (0, 6) are the endpoints of the diagonal of a square. What is the distance between point (0, 0) and the closest vertex of the square?

(A) 1/sqrt (2)
(B) 1
(C) sqrt (2)
(D) sqrt (3)
(E) 2*sqrt (3)

Can anyone explain with proper reasoning. I feel that the closest vertex should be 2 points away, so the distance should be 2

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Re: On the coordinate plane (6, 2) and (0, 6) are the endpoints [#permalink]

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24 Jul 2012, 13:39
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Expert's post
This is probably a question that is unrepresentative of the GMAT
however I will try to give you the shortest ans possible(Note that even the shortest ans requires you to solve a second degree equation)
As we can see one diagonal has the coordinates as (0,6) and (6,2)
So the slope of the diagonal is -2/3
As we know that the diagonals of a square bisect each other perpendicularly. the slope of the other diagonal is 3/2
Also the midpoint of the given diagonal is also the midpoint of the other one which is (3,4)
So, if we assume one end of the other diagonal to be (h,k), the other end of the diagonal becomes (6-h,8-k) with the slope being 3/2
Hereby we can find out that 3h-2k=1
Again we can see that the distance between (0,6) and (3,4) is the same as that between (h,k) and (3,4)
Solving a nasty equation we get, h=1,5 and from the equation one line earlier, we see k=1,7
So the vertices of the square are(0,6), (5,7), (6,2) and (1,1)
Clearly the vertex closest to Origin (0,0) is (1,1)
and in that case the distance between them is \sqrt{2}
Hence the ans is C
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Re: On the coordinate plane (6, 2) and (0, 6) are the endpoints [#permalink]

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25 Jul 2012, 08:31
Smita04 wrote:
On the coordinate plane (6, 2) and (0, 6) are the endpoints of the diagonal of a square. What is the distance between point (0, 0) and the closest vertex of the square?

(A) 1/sqrt (2)
(B) 1
(C) sqrt (2)
(D) sqrt (3)
(E) 2*sqrt (3)

Can anyone explain with proper reasoning. I feel that the closest vertex should be 2 points away, so the distance should be 2

Discussed here: coordinate-plane-90772.html

Check this solution: coordinate-plane-90772-60.html#p692465

Hope it helps.
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Re: On the coordinate plane (6, 2) and (0, 6) are the endpoints   [#permalink] 25 Jul 2012, 08:31
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