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# In the rectangular coordinate system, are the points (r,s)

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Senior Manager
Joined: 20 Feb 2006
Posts: 373
In the rectangular coordinate system, are the points (r,s) [#permalink]

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21 Feb 2006, 08:31
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100% (01:49) correct 0% (00:00) wrong based on 12 sessions

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In the rectangular coordinate system, are the points (r,s) and (u,v) equidistant from the origin?

1) r+s = 1

2) u = 1 - r, v = 1 - s
Senior Manager
Joined: 11 Jan 2006
Posts: 269
Location: Chennai,India

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21 Feb 2006, 08:35
it is B ,

1) we only know the value of r & s - insuffi

2) we know the value of u & s and can calculate r & s , since the distance from origin is asked - the calculation is simple.. - suffi
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Senior Manager
Joined: 20 Feb 2006
Posts: 373

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21 Feb 2006, 08:46
That's what I got but GMATPrep says c)??????
Intern
Joined: 13 Nov 2005
Posts: 9

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21 Feb 2006, 09:03
I think it is C.

The reason is :
With either assumption we cannot get an answer. Both have to be taken together to get the values we require.

Way2go
Manager
Joined: 13 Dec 2005
Posts: 224
Location: Milwaukee,WI

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21 Feb 2006, 09:41
The question here is r they equidistant or Not ... so the answer has to be YES OR NO .....

this u can only find by taking both A & B together .

by taking both the equation together u will get

u =s and v =r so the two coordinates turns out to be (r,s) and (s,r)

so both of them are equidistant from the origin.
Senior Manager
Joined: 20 Feb 2006
Posts: 373

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21 Feb 2006, 13:44
But will 2) not give us the information we need since for any value of r,s we know the corresponding value of u,v?
Manager
Joined: 13 Dec 2005
Posts: 224
Location: Milwaukee,WI

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21 Feb 2006, 14:22

going by what u said ... like choosing the option 2 and knowing the values of parameter u can findout whether its equidistant or not ... well here u dont know the values so if i ask u whether its equidistant or not u cant make out yes or no .

for example in take value r =1/2 ,s =1/2 so u =1/2 and v =1/2 ... means they are the same point and hence equidistant , now u put

r =2 ,s =2 so u =-1 and v = -1 so not equidistant ... so u really cannot
generalise from it what the answer will be ..definite yes or definite no

but when u take both 1 and 2 u get that irrespective of the value of the parameters, they are equi distant . Hence C .
Senior Manager
Joined: 20 Feb 2006
Posts: 373

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21 Feb 2006, 15:39
Thanks IPC,

I see what you're saying.
Intern
Joined: 31 May 2005
Posts: 12

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28 Feb 2006, 19:54
C

sqrt (r^2 +s^2) - sqrt (u^2+v^2) = sqrt ((1-r)^2 = (1-s)^2))

solving this by substituting given in 1) we arrive at C.
SVP
Joined: 14 Dec 2004
Posts: 1689

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01 Mar 2006, 11:49
This is "C".

we get
v^2+u^2 = r^2 + s^2 + 2 - 2(r+s)

Putting r+s = 1, we get, v^2+u^2 = r^2 + s^2
Intern
Joined: 17 Feb 2006
Posts: 11

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01 Mar 2006, 12:08
In the rectangular coordinate system, are the points (r,s) and (u,v) equidistant from the origin?

1) r+s = 1

2) u = 1 - r, v = 1 - s

1) gives us r+s but no information on u or v --> insufficient

2) gives us the distance from the origin as follows:

root [(1-r)^2 + (1-s)^2] = root [1 -2r + r^2 + 1 -2s + s^2]

= root [r^2 + s^2 + 2 - 2(r+s)]

From here we cannot tell whether the points are equidistant or not - it depends on whether the expression 2-2(r+s) = 0

Taking both (1) and (2) into account 2-2(r+s) = 2-2*1 = 0 --> sufficient

Re: Rectangular coordinate DS   [#permalink] 01 Mar 2006, 12:08
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