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In the xyplane, the points (c, d), (c, d), and (c, d) [#permalink]
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23 Jan 2012, 12:44
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In the xyplane, the points (c, d), (c, d), and (c, d) are three vertices of a certain square. If c < 0 and d > 0, which of the following points is in the same quadrant as the fourth vertex of the square? A. (5, 3) B. (5, 3) C. (5, 3) D. (3, 5) E. (3, 5) I answered B. But OA
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Re: In the xyplane, the points (c, d), (c, d), and (c, d) [#permalink]
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23 Jan 2012, 14:57
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Hi, there. I'm happy to help with this. The question: In the xyplane, the points (c, d), (c, d), and (c, d) are three vertices of a certain square. If c < 0 and d > 0, which of the following points is in the same quadrant as the fourth vertex of the square?I marked the tricky part in red. It seems c is a negative number and d is a positive number. This means Vertex #1 = (c, d) is in QII (that is, negative x and positive y) Vertex #2 = (c, d) is in QIII (that is, both x & y negative) Vertex #3 = (c, d) is in QIV (that is y is negative, but x is positive) That means the last vertex should be in the first quadrant  the only first quadrant point is (5, 3), answer = E. Does that make sense? Please let me know if you have any questions on this. Mike
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In the xyplane, the points (c, d), (c, d), and (c, d) [#permalink]
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23 Jan 2012, 17:16
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Baten80 wrote: In the xyplane, the points (c, d), (c, d), and (c, d) are three vertices of a certain square. If c < 0 and d > 0, which of the following points is in the same quadrant as the fourth vertex of the square?
A. (5, 3) B. (5, 3) C. (5, 3) D. (3, 5) E. (3, 5)
I answered B. But OA As the three vertices are (c, d), (c, d), and (c, d) then the fourth will be (c, d), the only combination which is left. Now, as c < 0 and d > 0, then \(c=negative=positive\) and \(d=positive\). Thus the fourth vertex \((c, d)=(positive, \ positive)\) is in the I quadrant, only answer choice E offers I quadrant point: (3, 5). Answer: E.
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Re: In the xyplane, the points (c, d), (c, d), and (c, d) [#permalink]
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24 Jan 2012, 02:04
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Baten80 wrote: In the xyplane, the points (c, d), (c, d), and (c, d) are three vertices of a certain square. If c < 0 and d > 0, which of the following points is in the same quadrant as the fourth vertex of the square?
A. (5, 3) B. (5, 3) C. (5, 3) D. (3, 5) E. (3, 5)
I answered B. But OA Your logic was fine.. You figured that after (c, d), (c, d), and (c, d), the fourth vertex would be (c, d). Perfect! The only problem was the additional info that said that 'c' is negative and 'd' is positive. This means that in (c, d), both (c) and d will be positive! Hence the OA is (E) i.e. the one in which both, the x and the y coordinate, are positive.
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In the xyplane, the points (c, d), (c, d), and (c, d) [#permalink]
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26 Oct 2016, 10:29
From the question since this is given If c < 0 and d > 0 .... you can do c = 1 and d = 1
Then map the coordinates for all three points (c,d) = (1, 1) , (c, d) = (1, 1) and (c, d) = (1,1) .....
Well if you draw all these three points in the plane you will realise that the missing point has positive x, y so E



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Re: In the xyplane, the points (c, d), (c, d), and (c, d) [#permalink]
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26 Jul 2017, 02:24
Very Tricky Question. Thanks for the explanations Mike/Bunuel.



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Re: In the xyplane, the points (c, d), (c, d), and (c, d) [#permalink]
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26 Jul 2017, 15:17
Baten80 wrote: In the xyplane, the points (c, d), (c, d), and (c, d) are three vertices of a certain square. If c < 0 and d > 0, which of the following points is in the same quadrant as the fourth vertex of the square?
A. (5, 3) B. (5, 3) C. (5, 3) D. (3, 5) E. (3, 5) Since c < 0 and d > 0: (c, d) = (neg, pos) = quadrant II (c, d) = (neg, neg) = quadrant III (c, d) = (pos, neg) = quadrant IV Thus, the 4th vertex is in quadrant I and has a point that is (pos, pos). Thus, choice E is correct, since it’s the only point that is (pos, pos). Answer: E
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Re: In the xyplane, the points (c, d), (c, d), and (c, d) [#permalink]
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30 Jul 2017, 09:51
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Although I understood the problem but is it possible for someone who can draw the vertices in a picture and present it. Visually still it is difficult for me to understand, how the c,d simply lying in the quadrant II.
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Re: In the xyplane, the points (c, d), (c, d), and (c, d) [#permalink]
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30 Jul 2017, 10:10



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Re: In the xyplane, the points (c, d), (c, d), and (c, d) [#permalink]
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23 Aug 2017, 14:46
Baten80 wrote: In the xyplane, the points (c, d), (c, d), and (c, d) are three vertices of a certain square. If c < 0 and d > 0, which of the following points is in the same quadrant as the fourth vertex of the square?
A. (5, 3) B. (5, 3) C. (5, 3) D. (3, 5) E. (3, 5)
I answered B. But OA I almost answered B, because it would match the quadrant; however, when you look at the final piece to the pattern; that is, (c, d) you realize 5 can not be c, because (5) = 5, which contradicts the given "c<0" After understanding what the question asks, The first thing you should do is list the givens: (c,d) (c,d) (c,d) (???) missing fourth point not given. Analyzing the points given, you see a pattern of c and d both positive, one positive and one negative, and both negative, the only pattern left is one negative and one positive. (c,d) is already given; therefore the missing fourth point is (c, d) (c,d) cannot be B, because, again, when you plug 5 into (c,d) you get ((5), d) which contradicts "c<0" Now, B and A are both eliminated. You know it cannot be an answer where d is a negative number, so C and D are also eliminated.Therefore, the only remaining answer is EBottomline: You should be able to recognize the pattern, and thereafter eliminate answer choices .



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Re: In the xyplane, the points (c, d), (c, d), and (c, d) [#permalink]
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13 Feb 2018, 20:51
Hi All, You might find it helpful to TEST VALUES. We're told that C < 0 and D > 0. Let's TEST C = 2 D = +2 We're given 3 of 4 coordinates for a square.... (C,D) = (2, +2) which is in the 2nd Quadrant (C, D) = (2, 2) which is in the 3rd Quadrant (C, D) = (+2, 2) which is in the 4th Quadrant Since we're dealing with a square, the 4th coordinate will be in 1st Quadrant...There's only one answer that's also in the 1st Quadrant... Final Answer: GMAT assassins aren't born, they're made, Rich
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