The graph of the equation xy = k, where k

Question

https://gmatclub.com/forum/download/file.php?id=28104 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 2015-10-16_0900.png

AkshdeepS · Accepted Answer

My Solution:

If k < 0 then the product of x and y = -k

It is only possible when either x or y is negative.

If x is negative and y positive (-x,y) = Quadrant II

If x is positive and y negative (x,-y) = Quadrant IV

Answer D

GMATinsight · Answer

To answer any such question, the easiest way is to choose the value of unknowns as per the given constraint (e.g. k = -10 or -20 etc.) and then substitute value of one of the variables and calculate the value of other variable as per the given expression (xy=k in this case) and finally plot them to answer the question

Please refer the figure below with assumed value of k = -10

Answer: option D

camlan1990 · Answer

x, y will be different in SIGN, meaning that x <0, y>0 or x>0, y<0 => Ans: D

Skywalker18 · Answer

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)

Answer D

JeffTargetTestPrep · Answer

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.

Answer: D

LogicGuru1 · Answer

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.
ANSWER IS D

mcelroytutoring · Answer

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.

unraveled · Answer

Graphical solution. Since xy < 0, either x or y is negative. xy = k.png Answer D.

avigutman · Answer

Video solution from Quant Reasoning: Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1 https://www.youtube.com/watch?v=rA8iYqpzYLc

GMATinsight · Answer

Answer: Option D

Video solution by  GMATinsight

https://www.youtube.com/watch?v=FJCyxCS3fLc

khera · Answer

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.

ArnauG · Answer

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).

GiovanniLERRO · Answer

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=  . 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.

kop18 · Answer

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?

Bunuel · Answer

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: https://gmatclub.com/forum/download/file.php?id=71824 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. Answer: D. desmos-graph (2).png

kop18 · Answer

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.

Answer: D.

Show Spoiler

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?

ScottTargetTestPrep · Answer

It's definitely not accurate to say that xy = k defines a line with equation y = k/x + 0. The equation of a line cannot have an x in the denominator. Since k is negative, point (0, 0) is not on the curve defined by the equation xy = k either. As others have pointed out, this equation defines a hyperbola, but you don't need to know that, and as far as I know, hyperbolas are not tested in GMAT either. The only observation you need to make to answer this question is that if the product of two numbers is negative, then one number must be positive and the other must be negative.

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

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