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# The graph of the equation xy = k, where k < 0, lies in which two of th

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Re: The graph of the equation xy = k, where k < 0, lies in which two of th [#permalink]
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Bunuel wrote:

The graph of the equation xy = k, where k < 0, lies in which two of the quadrants shown above?

(A) I and II
(B) I and III
(C) II and III
(D) II and IV
(E) III and IV

Kudos for a correct solution.

Attachment:
2015-10-16_0900.png

x, y will be different in SIGN, meaning that x <0, y>0 or x>0, y<0

=> Ans: D
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Re: The graph of the equation xy = k, where k < 0, lies in which two of th [#permalink]
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xy = k
and k <0
=> x< 0 or y <0
So ,the coordinates will have opposite signs .

Quadrant 2 : (-x , y )
Quadrant 4 : (x,-y)

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Re: The graph of the equation xy = k, where k < 0, lies in which two of th [#permalink]
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Bunuel wrote:

The graph of the equation xy = k, where k < 0, lies in which two of the quadrants shown above?

(A) I and II
(B) I and III
(C) II and III
(D) II and IV
(E) III and IV

Kudos for a correct solution.

Attachment:
2015-10-16_0900.png

Solution:

We are given the equation xy = k, where k is negative. It follows that x and y must have opposite signs. That is, either x is positive and y is negative, which is only possible in Quadrant IV, or, x is negative and y is positive, which is only possible in Quadrant II.

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Re: The graph of the equation xy = k, where k < 0, lies in which two of th [#permalink]
2
Kudos
Bunuel wrote:

The graph of the equation xy = k, where k < 0, lies in which two of the quadrants shown above?

(A) I and II
(B) I and III
(C) II and III
(D) II and IV
(E) III and IV

Kudos for a correct solution.

Attachment:
2015-10-16_0900.png

xy=k ==>xy=k+0
y=k/x +0
equation for line is y=mx+b
So we know y intercept is 0 which means when x=0, y=0
Meaning the line is passing through the origin (0,0) ==> perfect.
Now we need to figure out the slope==> k is the slope==> it is negative (given in question stem)==> line starts in Q II passes through the origin (0,0) and enters Q IV

Answer should have Q II and Q IV in it.
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Re: The graph of the equation xy = k, where k < 0, lies in which two of th [#permalink]
Top Contributor
Attached is a visual that should help. I suggest picking a simple negative number for k, such as -1, to make the concept more clear.

Note that the question doesn't say what k equals, so you are allowed to set the value of k to any negative number, and it should still work.
Attachments

Screen Shot 2017-05-30 at 4.07.06 PM.png [ 76.29 KiB | Viewed 48746 times ]

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Re: The graph of the equation xy = k, where k < 0, lies in which two of th [#permalink]
2
Kudos
Bunuel wrote:

The graph of the equation xy = k, where k < 0, lies in which two of the quadrants shown above?

(A) I and II
(B) I and III
(C) II and III
(D) II and IV
(E) III and IV

Kudos for a correct solution.

Attachment:
The attachment 2015-10-16_0900.png is no longer available

Graphical solution. Since xy < 0, either x or y is negative.
Attachment:

xy = k.png [ 41.89 KiB | Viewed 29473 times ]

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Re: The graph of the equation xy = k, where k < 0, lies in which two of th [#permalink]
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Video solution from Quant Reasoning:
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1
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Re: The graph of the equation xy = k, where k < 0, lies in which two of th [#permalink]
Bunuel wrote:

The graph of the equation xy = k, where k < 0, lies in which two of the quadrants shown above?

(A) I and II
(B) I and III
(C) II and III
(D) II and IV
(E) III and IV

Kudos for a correct solution.

Attachment:
2015-10-16_0900.png

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Re: The graph of the equation xy = k, where k < 0, lies in which two of th [#permalink]
If a point lies on the graph of xy = k,
then the
product of the point's x- and y-coordinates is k.
Since k is negative, it follows that for any such point,
the product of the point's x- and y-coordinates is
negative.
Therefore, for any such pbint, the point's
x-and y-coordinates have opposite signs, and hence
the point must be in quadrant II or in quadrant IV.
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Re: The graph of the equation xy = k, where k < 0, lies in which two of th [#permalink]
To determine in which two quadrants the graph of the equation xy = k lies, where k < 0, we can analyze the behavior of the equation in different quadrants.

Let's consider the signs of x and y in each quadrant:

Quadrant I: Both x and y are positive.
Quadrant II: x is negative, but y is positive.
Quadrant III: Both x and y are negative.
Quadrant IV: x is positive, but y is negative.
Given that k < 0, the equation xy = k implies that x and y have opposite signs. One of them is negative, while the other is positive.

From the above analysis, we can conclude that the graph of the equation xy = k lies in Quadrants II and IV (D).
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Re: The graph of the equation xy = k, where k < 0, lies in which two of th [#permalink]
You could solve this in under 30 seconds, you should be just aware of the different equation types and their respective graphs.
Organize in the correct form x*y=k, so that y=[k][/x]. This equation is a Hyperbole, in the most accessible form it is Y=1/x which exists in the I and III Q.
Now, just as quickly, Y=(-)k/x is the same hyperbole, BUT switching the Q, such that it exists in the II and IV Q.
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Re: The graph of the equation xy = k, where k < 0, lies in which two of th [#permalink]
LogicGuru1 wrote:
Bunuel wrote:

The graph of the equation xy = k, where k < 0, lies in which two of the quadrants shown above?

(A) I and II
(B) I and III
(C) II and III
(D) II and IV
(E) III and IV

Kudos for a correct solution.

Attachment:
2015-10-16_0900.png

xy=k ==>xy=k+0
y=k/x +0
equation for line is y=mx+b
So we know y intercept is 0 which means when x=0, y=0
Meaning the line is passing through the origin (0,0) ==> perfect.
Now we need to figure out the slope==> k is the slope==> it is negative (given in question stem)==> line starts in Q II passes through the origin (0,0) and enters Q IV

Answer should have Q II and Q IV in it.

Thank you for your solution.

I have a quick doubt - Please can you explain if it okay to have the equation of the line in the form y=k/x+0. I understand that the value of x (in any line y=mx+b) can be an integer or fraction but to have x coordinate in the denominator does not align with the slope equation. In the slope intercept form x is multiplied with the slope and not divided. Hence, is it not different to form a line that has a fractional value of x from having a line that has the x cooridnate in the denominator?

In simpler terms is it okay to have a slope equation of the form that is being made here?

I am not sure if there's conceptual gap in my understanding? Please can mike GMATNinja ian KarishmaB marty scott anyone help explain this to me?
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Re: The graph of the equation xy = k, where k < 0, lies in which two of th [#permalink]
1
Kudos
kop18 wrote:
LogicGuru1 wrote:
Bunuel wrote:

The graph of the equation xy = k, where k < 0, lies in which two of the quadrants shown above?

(A) I and II
(B) I and III
(C) II and III
(D) II and IV
(E) III and IV

Kudos for a correct solution.

Attachment:
The attachment 2015-10-16_0900.png is no longer available

xy=k ==>xy=k+0
y=k/x +0
equation for line is y=mx+b
So we know y intercept is 0 which means when x=0, y=0
Meaning the line is passing through the origin (0,0) ==> perfect.
Now we need to figure out the slope==> k is the slope==> it is negative (given in question stem)==> line starts in Q II passes through the origin (0,0) and enters Q IV

Answer should have Q II and Q IV in it.

Thank you for your solution.

I have a quick doubt - Please can you explain if it okay to have the equation of the line in the form y=k/x+0. I understand that the value of x (in any line y=mx+b) can be an integer or fraction but to have x coordinate in the denominator does not align with the slope equation. In the slope intercept form x is multiplied with the slope and not divided. Hence, is it not different to form a line that has a fractional value of x from having a line that has the x cooridnate in the denominator?

In simpler terms is it okay to have a slope equation of the form that is being made here?

I am not sure if there's conceptual gap in my understanding? Please can mike GMATNinja ian KarishmaB marty scott anyone help explain this to me?

I think you're under the impression that xy = k is the equation of a line, which is not accurate. In fact, xy = k, where k ≠ 0, results in a hyperbola. For instance, see this graph of xy = -10:

To solve this question, you don't really need all of this. Having xy equal to a negative number means that when x is positive, y is negative. Therefore, in this case, all points are in the IV quadrant. Conversely, when x is negative, y is positive. Thus, in this case, all points are in the II quadrant.

Attachment:

desmos-graph (2).png [ 60.96 KiB | Viewed 5009 times ]
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Re: The graph of the equation xy = k, where k < 0, lies in which two of th [#permalink]
Thank you for your solution.

I have a quick doubt - Please can you explain if it okay to have the equation of the line in the form y=k/x+0. I understand that the value of x (in any line y=mx+b) can be an integer or fraction but to have x coordinate in the denominator does not align with the slope equation. In the slope intercept form x is multiplied with the slope and not divided. Hence, is it not different to form a line that has a fractional value of x from having a line that has the x cooridnate in the denominator?

In simpler terms is it okay to have a slope equation of the form that is being made here?

I am not sure if there's conceptual gap in my understanding? Please can mike GMATNinja ian KarishmaB marty scott anyone help explain this to me?[/quote]

I think you're under the impression that xy = k is the equation of a line, which is not accurate. In fact, xy = k, where k ≠ 0, results in a hyperbola. For instance, see this graph of xy = -10:

To solve this question, you don't really need all of this. Having xy equal to a negative number means that when x is positive, y is negative. Therefore, in this case, all points are in the IV quadrant. Conversely, when x is negative, y is positive. Thus, in this case, all points are in the II quadrant.

Attachment:
desmos-graph (2).png
[/quote]

Thanks Bunuel this makes sense. I got a little confused with the solution above. This clarifies my doubt as to why the equation in this question is not that of a line and actually forms a parabola. Hence, will not be of the form y=mx+b.

Just an add on question - Can you confirm if there can be a case/possibility to have an equation (in which the x coordinate is in the denominator) similar to the one that is formed in this question?
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Re: The graph of the equation xy = k, where k < 0, lies in which two of th [#permalink]