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

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

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New post 15 Oct 2015, 22:01
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A
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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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New post 16 Oct 2015, 00:12
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Bunuel wrote:
Image
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


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Attachment:
2015-10-16_0900.png


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

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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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New post 15 Oct 2015, 23:41
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Bunuel wrote:
Image
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


y=k/x (0,0); when x is +ve y will be negative and vice versa --> Quadrant 2 and 4 Answer (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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New post 16 Oct 2015, 01:37
2
Bunuel wrote:
Image
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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New post 16 Oct 2015, 02:45
2
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
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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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New post 16 Oct 2015, 08:02
1
XY=K and K<0 ; so we can say that
1. X is negative y is positive
2. X is positive y is negative
3. neither of them will not be zero

Hence to satisfy above conditions; equation xy<o has to be in 2nd and 4th quadrant where one of them will be positive and another will be negative.
Hence answer is 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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New post 17 Oct 2015, 04:50
1
Bunuel wrote:
Image
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


D - If you the graph of y=1/x then just take the reflection of that curve about the axis to get y=-1/x
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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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New post 17 Oct 2015, 07:51
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Bunuel wrote:
Image
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


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

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New post 13 Feb 2016, 18:34
Can this equation be brought into the y=mx+b form or is it a quadratic. If it can then what is the slope for this equation considering it has two different lines in different quadrants?

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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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New post 11 May 2016, 02:03
Bunuel wrote:
Image
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

We know xy<0.
So x&y will have opposite signs. Now in Q3 & Q1, x&y have equal signs. So only possible placement for line will be out of Q1 and Q4.
That means line xy=k will lie on Q3&Q4.

So answer is 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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New post 11 May 2016, 07:22
1
Bunuel wrote:
Image
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.

Answer: 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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New post 14 Jul 2016, 05:51
Bunuel wrote:
Image
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.
ANSWER IS 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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New post 17 Nov 2016, 09:41
Bunuel wrote:
Image
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

here,
K=negative
xy=k
xy=(-)
(-) (+)=(-) ----->which indicates quadrant II
(+) (-)=(-) ----->which indicates quadrant IV

So, Correct answer is D
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The graph of the equation xy = k, where k < 0, lies in which two of th  [#permalink]

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New post 30 May 2017, 16:11
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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.
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Screen Shot 2017-05-30 at 4.07.06 PM.png [ 76.29 KiB | Viewed 16381 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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New post 04 Jun 2017, 01:20
If k < 0 then the product of x and y = -k

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

x is -ve & y +ve (-x,y) = Quadrant II

Option should be A, C, D

Lets look at point second condition
x is +veand y -ve (x,-y) = Quadrant IV

Ans D
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Re: The graph of the equation xy = k, where k < 0, lies in which two of th   [#permalink] 04 Jun 2017, 01:20
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