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

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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) are three ve  [#permalink]

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

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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)

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).

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

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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)

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) are three ve  [#permalink]

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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) are three ve  [#permalink]

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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) are three ve  [#permalink]

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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)

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) are three ve  [#permalink]

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

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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 10791 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) are three ve  [#permalink]

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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)

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 .
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GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: In the xy-plane, the points (c, d), (c, -d), and (-c, -d) are three ve  [#permalink]

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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) are three ve  [#permalink]

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Simply test values for c = -1 and d = 1 (per constraints of the question)
Since all the sides are equal, the point must lie in the first quadrant and EVERY point in this quadrant is in the form (x,y) where x>0 and y > 0.

E is the only answer choice that aligns.
Attachments Capture.JPG [ 95.9 KiB | Viewed 2162 times ]

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

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mikemcgarry wrote:
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 Hi,

I am a bit confused here to understand how did you come across, Vertex #2 = (c, -d) is in QIII (that is, both x & y negative) while you are saying Vertex #1 = (c, d) is in QII (that is, negative x and positive y)

Because, for D to become negative, the original value of D should be a positive value with the condition given d > 0.

So do we have to consider the original value of the points or the value after multiplying +/-?

Can you please explain this.

mihiri
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GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: In the xy-plane, the points (c, d), (c, -d), and (-c, -d) are three ve  [#permalink]

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Hi Mihiri,

We're told that C < 0 and D > 0... so before you do anything to those values (such as multiply them by "-1"), you have to make sure that you make sure that your C and D fit those facts first.

To help you with how all of this applies to this question, you might find it helpful to TEST VALUES.

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) are three ve   [#permalink] 22 Aug 2019, 14:39
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