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# In the xy plane point P (m,n) and point Q (n,m) what is the

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In the xy plane point P (m,n) and point Q (n,m) what is the [#permalink]  22 Sep 2010, 12:07
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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
[Reveal] Spoiler: OA

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Re: Co-ordinate geometry DS [#permalink]  22 Sep 2010, 12:13
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rxs0005 wrote:
In the xy plane point P ( m,n) and point Q ( n,m) what is the distance between P and Q

m - n = 2

m + n = 5

The formula to calculate the distance between two points $$(x_1,y_1)$$ and $$(x_2,y_2)$$ is $$d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$$.

So we are asked to calculate: $$d=\sqrt{(m-n)^2+(m-n)^2}=\sqrt{2(m-n)^2}=\sqrt{2}*|m-n|$$.

(1) $$m-n=2$$. Sufficient.

(2) $$m+n=5$$. Not sufficient.

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Re: Co-ordinate geometry DS [#permalink]  22 Sep 2010, 12:14
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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: Co-ordinate geometry DS [#permalink]  22 Sep 2010, 12:17
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Re: Co-ordinate geometry DS [#permalink]  22 Sep 2010, 12:45
Thanks Gentlemen for the algebraic approach i was using the graphical approach and was confused
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Re: Co-ordinate geometry DS [#permalink]  22 Sep 2010, 13:02
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: Co-ordinate geometry DS [#permalink]  22 Sep 2010, 13:25
rxs0005 wrote:
Thanks Gentlemen for the algebraic approach i was using the graphical approach and was confused

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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Re: Co-ordinate geometry DS [#permalink]  22 Sep 2010, 13:29
shrouded1 wrote:
rxs0005 wrote:
Thanks Gentlemen for the algebraic approach i was using the graphical approach and was confused

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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Re: Co-ordinate geometry DS [#permalink]  22 Sep 2010, 13:31
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: Co-ordinate geometry DS [#permalink]  22 Sep 2010, 13:33
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: Co-ordinate geometry DS [#permalink]  29 Oct 2010, 17:47
+1 A
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Re: Co-ordinate geometry DS [#permalink]  09 May 2013, 18:20
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.
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Re: Co-ordinate geometry DS [#permalink]  09 May 2013, 20:46
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josemnz83 wrote:
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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Re: In the xy plane point P (m,n) and point Q (n,m) what is the [#permalink]  14 Nov 2013, 03:09
I solved this question by drawing a rough diagram.

Option A > m-n = 2 which means OP = 2, OQ =2

Since OPQ is a right triangle, we can easily calculate the distance PQ. Sufficient

Option B > m+n = 2, clearly insufficient to come to any solution

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Re: In the xy plane point P (m,n) and point Q (n,m) what is the   [#permalink] 14 Nov 2013, 03:09
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