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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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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 (m-n) & its diagnol is the straight line joining (m,n) and (n,m)

So distance = Sqrt(2) * (m-n)

Hence all you need is m-n

The solution seems a bit involved, but if you draw it out it'll be pretty straight forward.
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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 (m-n) & its diagnol is the straight line joining (m,n) and (n,m)

So distance = Sqrt(2) * (m-n)

Hence all you need is m-n

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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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

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

I'm not understanding how A is sufficient to answer the question. All A says is that m-n=2. What about n-m? How are m-n and n-m 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.

(m-n) and (n-m) are not the same thing, however, (m-n)^2 and (n-m)^2 are the same thing. Take (m-n) = 2, thus (n-m) = -2. However, (m-n)^2 = (n-m)^2 = 4.
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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 = \([(m-n)^2 + (n-m)^2]\)^(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, 09:09

The distance would be Square root of (m-n)^2 + (n-m)^2 therefore m-n = 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.

gmatclubot

Re: In the xy plane point P (m,n) and point Q (n,m) what is the
[#permalink]
07 Jun 2016, 09:09

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