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

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In the rectangular coordinate system, are the points (r,s) [#permalink] New post 15 Jun 2006, 00:12
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A
B
C
D
E

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(N/A)

Question Stats:

50% (02:07) correct 50% (00:21) wrong based on 1 sessions
In the rectangular coordinate system, are the points (r,s) and (u,v) equidistant from the origin?

condition 1: r +s = 1

condition 2: u = 1 -r and v = 1 - s

FYI: I got the answer by picking numbers. Can someone please explain the "correct" mathematically method for solving this problem? Specifically, how does the constraint given by condition 1 algebrically relate to condition 2? Thanks for your help.

Mike
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 [#permalink] New post 15 Jun 2006, 01:59
I get (C)

Equidistant from 0 is the meaning that the 2 points are on the same circle with the origine O.

The equation of circle from origine is : x^2 + y^2 = r^2

So, we want to know either r^2 + s^2 = u^2 + v^2

1) r + s = 1
INSUFF

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

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


(1) & (2) :
r + s = 1 thus,

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

The 2 points are equidistants

Last edited by Fig on 15 Jun 2006, 02:16, edited 2 times in total.
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Re: DS: Geometry (a difficult problem i believe) [#permalink] New post 15 Jun 2006, 02:09
mrmikec wrote:
In the rectangular coordinate system, are the points (r,s) and (u,v) equidistant from the origin?

condition 1: r +s = 1

condition 2: u = 1 -r and v = 1 - s

FYI: I got the answer by picking numbers. Can someone please explain the "correct" mathematically method for solving this problem? Specifically, how does the constraint given by condition 1 algebrically relate to condition 2? Thanks for your help.

Mike


For points to be equidistant from origin (0,0)

Sqrt(r^2 + s^2) = Sqrt(u^2 + v^2)
r^2 + s^2 =u^2 + v^2
s^2-v^2 = u^2-r^2
(s-v)(s+v) = (u-r)(u+r) --- A

Statement 1 - Insufficient as says nothing about u & v

Statement 2
Helps us reduce condition A to
(s-v) = (u-r)
or
s+r = u+v
= 2 -(r+s)
r+s = 1 --- this condition is still not satisfied by Statement 2 - Hence Insuff

Together sufficient as we get r+s = 1 from Statement 1
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 [#permalink] New post 15 Jun 2006, 04:47
Agree with Shobhitb and Fig. It is C. Great explanations
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 [#permalink] New post 15 Jun 2006, 13:04
Fig wrote:
I get (C)

Equidistant from 0 is the meaning that the 2 points are on the same circle with the origine O.

The equation of circle from origine is : x^2 + y^2 = r^2

So, we want to know either r^2 + s^2 = u^2 + v^2

1) r + s = 1
INSUFF

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

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


(1) & (2) :
r + s = 1 thus,

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

The 2 points are equidistants


This is an awesome explanation. Thanks for the effort.

I got to here

= 1 - 2*r + r^2 + 1 - 2*s + s^2
= 2 - 2*(r+s) + r^2 + s^2

when i did the problem. i didn't think to factor out the 2r + 2s to sub in the r+s =1. I always seem to forgot to sub back an equation stem from condition 1 or the question stem. did thoughts on how to remedy that? (besides more practice which i'm doing.
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 [#permalink] New post 16 Jun 2006, 12:14
Ohhh yes :) I understand what u are looking for :)

Unfortunatly, I have no special ways that offer a systematic good manner to answer or build the answer of such DS problems ;)

I would like perhaps to suggest to always expect a relationship more or less direct and/or easy to guess. Since we are ready and we know that such relationships should exist, perhaps it's the little plus that makes the difference :)

Good luck ! :)
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 [#permalink] New post 16 Jun 2006, 12:41
C

Distance of point A from origin, A= r^2 + s^2
Distance of point B from origin, B = u^2 + v^2

St1: r+s = 1 : Clearly INSUFF

St2: We get
B = (1-r)^2 + (1-s)^2 = 1-2r+r^2+1-2s+s^2 i.e
B = 2-2*(r+s) + r^2 + s^2 : INSUFF

Combined:

Substitute St1 in above equation and we get

B = 2-2*1 + r^2 + s^2 = r^2 + s^2 i.e B = A
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SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

  [#permalink] 16 Jun 2006, 12:41
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