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In the xy plane point P (m,n) and point Q (n,m) what is the [#permalink]
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22 Sep 2010, 13:07
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In the xy plane point P (m,n) and point Q (n,m) what is the distance between P and Q (1) m  n = 2 (2) m + n = 5
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Re: Coordinate geometry DS [#permalink]
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22 Sep 2010, 13:13



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Re: Coordinate geometry DS [#permalink]
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22 Sep 2010, 13:14
Distance between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(\sqrt{(x_1  x_2)^2 + (y_1  y_2)^2}\) put the values you will get \(\sqrt{(m  n)^2 + (n  m)^2}\) = \(\sqrt{2(m  n)^2}\) Hence A is sufficient.
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Re: Coordinate geometry DS [#permalink]
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22 Sep 2010, 13:17
I m late !!
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Re: Coordinate geometry DS [#permalink]
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22 Sep 2010, 13:45
Thanks Gentlemen for the algebraic approach i was using the graphical approach and was confused
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Re: Coordinate geometry DS [#permalink]
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22 Sep 2010, 14:02
A. You can not rely on the graphical solution. I mean you can plot it, but then eventually you will have to play with the equation for distance between two points.
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Re: Coordinate geometry DS [#permalink]
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22 Sep 2010, 14:25
rxs0005 wrote: Thanks Gentlemen for the algebraic approach i was using the graphical approach and was confused adishail wrote: A.
You can not rely on the graphical solution. I mean you can plot it, but then eventually you will have to play with the equation for distance between two points. Actually you can solve this graphically. What you need to note is that points (m,n) and (n,m) are mirror reflections across the x=y (45 degree) line on the coordinate plane.And the final piece in the puzzle is, if you draw a horizontal and a vertical line from (m,n) & the point (n,m) to intersect x=y, both lines will intersect at the points (m,m) and (n,n). This square you form has side (mn) & its diagnol is the straight line joining (m,n) and (n,m) So distance = Sqrt(2) * (mn) Hence all you need is mn The solution seems a bit involved, but if you draw it out it'll be pretty straight forward.
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Re: Coordinate geometry DS [#permalink]
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22 Sep 2010, 14:29
shrouded1 wrote: rxs0005 wrote: Thanks Gentlemen for the algebraic approach i was using the graphical approach and was confused adishail wrote: A.
You can not rely on the graphical solution. I mean you can plot it, but then eventually you will have to play with the equation for distance between two points. Actually you can solve this graphically. What you need to note is that points (m,n) and (n,m) are mirror reflections across the x=y (45 degree) line on the coordinate plane.And the final piece in the puzzle is, if you draw a horizontal and a vertical line from (m,n) & the point (n,m) to intersect x=y, both lines will intersect at the points (m,m) and (n,n). This square you form has side (mn) & its diagnol is the straight line joining (m,n) and (n,m) So distance = Sqrt(2) * (mn) Hence all you need is mn The solution seems a bit involved, but if you draw it out it'll be pretty straight forward. Actually, it is the same thing as plotting two points and then writing the equation of distance between two points. As far as being a faster way, the fact that the pyth theorem uses a difference is enough to get the soultion.
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Re: Coordinate geometry DS [#permalink]
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22 Sep 2010, 14:31
Ultimately it has to be the same thing, same problem same solution ... just saying that you can solve the question by plotting a graph and without knowing the distance formula Alternate solution if you may
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Re: Coordinate geometry DS [#permalink]
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22 Sep 2010, 14:33
shrouded1 wrote: Ultimately it has to be the same thing, same problem same solution ... just saying that you can solve the question by plotting a graph and without knowing the distance formula
Alternate solution if you may Agree
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Re: Coordinate geometry DS [#permalink]
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09 May 2013, 19:20
I'm not understanding how A is sufficient to answer the question. All A says is that mn=2. What about nm? How are mn and nm the same? gurpreetsingh wrote: Distance between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(\sqrt{(x_1  x_2)^2 + (y_1  y_2)^2}\)
put the values you will get \(\sqrt{(m  n)^2 + (n  m)^2}\)
= \(\sqrt{2(m  n)^2}\)
Hence A is sufficient.



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Re: Coordinate geometry DS [#permalink]
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09 May 2013, 21:46
josemnz83 wrote: I'm not understanding how A is sufficient to answer the question. All A says is that mn=2. What about nm? How are mn and nm the same? gurpreetsingh wrote: Distance between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(\sqrt{(x_1  x_2)^2 + (y_1  y_2)^2}\)
put the values you will get \(\sqrt{(m  n)^2 + (n  m)^2}\)
= \(\sqrt{2(m  n)^2}\)
Hence A is sufficient. (mn) and (nm) are not the same thing, however, (mn)^2 and (nm)^2 are the same thing. Take (mn) = 2, thus (nm) = 2. However, (mn)^2 = (nm)^2 = 4.
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Re: In the xy plane point P (m,n) and point Q (n,m) what is the [#permalink]
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14 Nov 2013, 04:09
I solved this question by drawing a rough diagram. Option A > mn = 2 which means OP = 2, OQ =2 Since OPQ is a right triangle, we can easily calculate the distance PQ. SufficientOption B > m+n = 2, clearly insufficient to come to any solution Correct answer A.
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In the xyplane, point P has coordinates (m; n) and point Q has coordi [#permalink]
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Updated on: 10 Apr 2016, 11:35
In the xyplane, point P has coordinates (m; n) and point Q has coordinates (n;m). What is the distance between P and Q? (1) m  n = 2 (2) m + n = 5 Hello,
I will be more than happy if you can help me with the explanation please as I do not see much difference between (1) and (2).
Originally posted by Jaikuz on 10 Apr 2016, 11:32.
Last edited by Bunuel on 10 Apr 2016, 11:35, edited 1 time in total.
Renamed the topic and edited the question.



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Re: In the xy plane point P (m,n) and point Q (n,m) what is the [#permalink]
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10 Apr 2016, 11:37



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In the xy plane point P (m,n) and point Q (n,m) what is the [#permalink]
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10 Apr 2016, 20:26
Jaikuz wrote: In the xyplane, point P has coordinates (m; n) and point Q has coordinates (n;m). What is the distance between P and Q? (1) m  n = 2 (2) m + n = 5 Hello,
I will be more than happy if you can help me with the explanation please as I do not see much difference between (1) and (2). There is a lot of difference between 1 and 2. The sign makes all the difference. Distance between two points \((x_1, y_1)\) and \((x_2, y_2)\) = \([(x_2  x_1)^2 (y_2  y_1)^2]\)^(1/2) In the given question, Distance = \([(mn)^2 + (nm)^2]\)^(1/2) Since this formula uses a  sign, we can solve this by using the statement 1, but not statement 2. Does this help?



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Re: In the xy plane point P (m,n) and point Q (n,m) what is the [#permalink]
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07 Jun 2016, 10:09
The distance would be Square root of (mn)^2 + (nm)^2 therefore mn = 5 can be substituted in this equation to obtain the actual distance. The second one gives value of m+n that cannot be utilized without further info in this equation hence A.



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Re: In the xy plane point P (m,n) and point Q (n,m) what is the [#permalink]
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Re: In the xy plane point P (m,n) and point Q (n,m) what is the
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