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

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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)

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

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23 Jan 2012, 15:57
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Hi, there. I'm happy to help with this.

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)  [#permalink]

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23 Jan 2012, 18:16
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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)

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).

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

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24 Jan 2012, 03:04
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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)

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

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26 Oct 2016, 11:29
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 xy-plane, the points (c, d), (c, -d), and (-c, -d)  [#permalink]

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26 Jul 2017, 03:24
Very Tricky Question. Thanks for the explanations Mike/Bunuel.
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Re: In the xy-plane, the points (c, d), (c, -d), and (-c, -d)  [#permalink]

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26 Jul 2017, 16:17
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)

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).

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

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30 Jul 2017, 11:10
1
AnubhavK wrote:
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 6680 times ]

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

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

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13 Feb 2018, 21:51
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 co-ordinates 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 co-ordinate will be in 1st Quadrant...There's only one answer that's also in the 1st Quadrant...

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

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08 Mar 2019, 04:29
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Re: In the xy-plane, the points (c, d), (c, -d), and (-c, -d)   [#permalink] 08 Mar 2019, 04:29
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# In the xy-plane, the points (c, d), (c, -d), and (-c, -d)

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