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In the xy-plane, the points (c, d), (c, -d), and (-c, -d) [#permalink]

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23 Jan 2012, 13:44

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In the xy-plane, 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)

The question: In the xy-plane, 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 xy-plane, 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 (positive, positive) is in the I quadrant, only answer choice E offers also I quadrant point: (3, 5).

In the xy-plane, 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 co-ordinate, are positive.
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Re: In the xy-plane, the points (c, d), (c, -d), and (-c, -d) [#permalink]

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06 Aug 2014, 05:03

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Re: In the xy-plane, the points (c, d), (c, -d), and (-c, -d) [#permalink]

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01 Sep 2015, 10:55

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Re: In the xy-plane, the points (c, d), (c, -d), and (-c, -d) [#permalink]

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07 Oct 2016, 08:01

Hello from the GMAT Club BumpBot!

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In the xy-plane, 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 xy-plane, the points (c, d), (c, -d), and (-c, -d) [#permalink]

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30 Jul 2017, 10:51

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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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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.

Check the diagram below:

Attachment:

Untitled.png [ 13.42 KiB | Viewed 1298 times ]

The fourth vertex is at (-c, d), which is (positive, positive), so in the first quadrant.

Re: In the xy-plane, the points (c, d), (c, -d), and (-c, -d) [#permalink]

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23 Aug 2017, 15:46

Baten80 wrote:

In the xy-plane, 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 E

Bottomline: You should be able to recognize the pattern, and thereafter eliminate answer choices .