Find all School-related info fast with the new School-Specific MBA Forum

It is currently 24 Aug 2016, 10:03
GMAT Club Tests

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

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

  post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
Director
Director
avatar
Joined: 03 Jul 2003
Posts: 652
Followers: 3

Kudos [?]: 73 [0], given: 0

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

Show Tags

New post 13 Dec 2003, 21:59
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

In the rectangular coordinate system, are the points (r, s) and (u, v) equidistant from the origin ?
(1) r + s = 1
(2) u = 1 тАУ r and v = 1 тАУ s
Director
Director
avatar
Joined: 03 Jul 2003
Posts: 652
Followers: 3

Kudos [?]: 73 [0], given: 0

 [#permalink]

Show Tags

New post 13 Dec 2003, 22:07
Explanation with steps please?
Director
Director
User avatar
Joined: 14 Oct 2003
Posts: 587
Location: On Vacation at My Crawford, Texas Ranch
Followers: 1

Kudos [?]: 39 [0], given: 0

 [#permalink]

Show Tags

New post 13 Dec 2003, 22:24
okay. we want to know whether pts (r,s) and (u,v) are equidistant from the origin (point 0,0) in the coordinate plane.

that means you can have values for r,s and u,v as (-1,0) and (1,0) or (-3,3) and (3,3) or (0,4) and (0,-4) etc etc. you get the picture

1. does r+s=1 tell us anything about point (u,v)? insufficient!

2. rewrite this to u+r=1 and v+s=1

okay few scenarios here

if (r,s) = (0,1) and (u,v) = (1,0) the two coordinates would be equidistant

however if you had values for (r,s) = (-2,-3) and (u,v) = (3,4) then they would NOT be equidistant.
Insufficient

However, they would be sufficient if you had 1 and 2.
no matter how what scenario you run through for both coordinates, with both 1 and 2 you'll have equidistant points.

try it out:

(r,s) = .5, .5 then (u,v) must equal .5 and .5
(r,s) = 0,1 then (u,v) must equal 1,0
(r,s) = 1,0 then (u,v) must equal 0,1
(r,s) = -1,2 then (u,v) must equal 2,-1

hope this helps
Director
Director
User avatar
Joined: 13 Nov 2003
Posts: 964
Location: Florida
Followers: 1

Kudos [?]: 92 [0], given: 0

 [#permalink]

Show Tags

New post 13 Dec 2003, 23:30
agree
when r+s = 1, then r,u and v,s will always have the opposite co-ordinates
SVP
SVP
User avatar
Joined: 30 Oct 2003
Posts: 1793
Location: NewJersey USA
Followers: 5

Kudos [?]: 81 [0], given: 0

 [#permalink]

Show Tags

New post 14 Dec 2003, 13:06
C is the answer. See the following explaination

let x1 be the distance from (0,0) to (r,s)
so
sqr(x1) = sqr(r) + sqr(s)

similarly x2 be the distance from (0,0) to (u,v)
so
sqr(x2) = sqr(u) + sqr(v)

Condition 1 does not say anything aout relation between (r,s) and (u,v)
so it cannot convey anything

Condition 2 however talks about relation between (r,s) and (u,v)

sqr(x2) = sqr(u) + sqr(v) = sqr(1-r) + sqr(1-s)

= 1 + sqr(r) - 2r + 1 + sqr(s) - 2s

= 1 + sqr(r) + sqr(s) - 2( r+s )
Again we cannot understand anything from this condition

-------------------------------------------------------------

if we combine condition 1 and 2
u + v = 2-(r+s) = 1 ( since r+s=1 from condition 1 )
so u+v = r+s

We can say sqr(u) + sqr(v) = sqr((u+v) - 2uv = 1-2uv ( which sqr of x2 )

Now we dont know the value of uv so we use second condition to
multiply u and v
uv = (1-r) * (1-s) = 1+rs - (r+s) = 1 + rs -1 = rs
therefore uv = rs

We can say sqr(r) + sqr(s) = sqr(r+s) - 2rv = 1-2sr = 1-2uv

Ultimatey x1 = x2 hence they are equidistant
  [#permalink] 14 Dec 2003, 13:06
Display posts from previous: Sort by

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

  post reply Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.