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

It is currently 22 May 2013, 10:12
Customize  |  Hide

In the rectangular coordinate system, are the points (a, b)

  Question banks Downloads My Bookmarks Reviews  
Author Message
TAGS:
Director
Director
Joined: 11 Jun 2007
Posts: 647
Followers: 1

Kudos [?]: 16 [0], given: 0

GMAT Tests User
In the rectangular coordinate system, are the points (a, b) [#permalink] New post 10 Dec 2007, 16:44
00:00

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
In the rectangular coordinate system, are the points (a, b) and (c, d) equidistant from the origin?

(1) a/b = c/d

(2) sqrt(a^2) + sqrt(b^2) = sqrt(c^2) + sqrt(d^2)
1 KUDOS received
SVP
SVP
User avatar
Joined: 01 May 2006
Posts: 1837
Followers: 8

Kudos [?]: 71 [1] , given: 0

GMAT Tests User
 [#permalink] New post 11 Dec 2007, 13:21
1
This post received
KUDOS
(C) too for me :)

Equidistant to 0 means:
sqrt(a^2 + b^2) = sqrt(c^2 + d^2)

In other words, the points A(a,b) and C(c,d) are both on a same circle, centered at (0,0).

Stat1
a/b = c/d = 1/s where s is a real number different from 0.

That implies:
o b = s*a
o d = s*c

Could we see an anology? :)... It's the equation of a line : y = s*x, passing by 0(0,0).

From here, we understand that, such points, can (sometimes) or cannot (most of time) be equidistant to 0.

o If a=-c and b=-d, we have the special case of equidistance. The line shares the points with a circle centered at 0(0,0)
o If a=c and d=d, both points are confused and so are equidistant to 0(0,0)
o All other cases are represented by the line y = s*x and do not create an equidistance (0 is out of the line ;))

INSUFF.

Stat2
o sqrt(a^2) + sqrt(b^2) = sqrt(c^2) + sqrt(d^2)
<=> |a| + |b| = |c| + |d|

We have 3 combinations:
o If |a| = |c|, then |b| = |d| that implies that A and B are in 4 possible points : the vertice of a square centered at 0(0,0)... (a,b), (-a,b), (a,-b) (-a,-b)
o If |a| = |d|, then |b| = |c| that implies that A and B are symetrical to the line y = x or on points rotated from 0(0,0) by k * 90° of these points (creating similar points in cadran 2, 3 and 4)... All of them are on a same circle centered at 0(0,0)
o All other cases : especially |b| = 0 and c!=0,d!=0... |a| = |c| + |d|

Concreate examples:
o If a=c=1 and b=d=0, then sqrt(a^2 + b^2) = sqrt(1+0)=1 and sqrt(c^2 + b^2) = sqrt(1+0) = 1.... equidistant to 0
o If a=2, b=0 and c=1 and d=1, then sqrt(a^2+b^2) = sqrt(2^2 + 0) = 2 and sqrt(c^2 + d^2) = sqrt(1 + 1) = sqrt(2).... not equidistant to 0

INSUFF.

Both 1 and 2
We have:
o b=s*a and d = s*c
o |a| + |b| = |c| + |d|

That implies:
o |a| + |s*a| = |c| + |s*c|
<=>(1+|s|)*|a| = (1+|s|)*|c|
<=> |a| = |c|

So,
o |a| + |b| = |c| + |d|
<=> |a| + |b| = |a| + |d|
<=> |b| = |d|

Bingo!... We are on the special case of:
o If |a| = |c|, then |b| = |d| that implies that A and B are in 4 possible points : the vertice of a square centered at 0(0,0)... (a,b), (-a,b), (a,-b) (-a,-b)

SUFF.
Intern
Intern
User avatar
Joined: 01 Dec 2007
Posts: 34
Followers: 0

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

 [#permalink] New post 10 Dec 2007, 16:56
I think it's C.

1) NS because it only tells us that the distances are proportional, e.g. 2, 3 : 4, 6..

2) NS because a = 2, b = 3, c=1, d=5 satisfies the condition but the points are not equidistant.

1&2) Sufficient
Director
Director
User avatar
Joined: 08 Jun 2007
Posts: 586
Followers: 1

Kudos [?]: 66 [0], given: 0

GMAT Tests User
 [#permalink] New post 10 Dec 2007, 17:22
applecrisp wrote:
I think it's C.

1) NS because it only tells us that the distances are proportional, e.g. 2, 3 : 4, 6..

2) NS because a = 2, b = 3, c=1, d=5 satisfies the condition but the points are not equidistant.

1&2) Sufficient


How do you prove that 1 and 2 together are sufficient.
I am getting E
Director
Director
Joined: 11 Jun 2007
Posts: 647
Followers: 1

Kudos [?]: 16 [0], given: 0

GMAT Tests User
 [#permalink] New post 11 Dec 2007, 14:07
Fig wrote:
(C) too for me :)

That implies:
o |a| + |s*a| = |c| + |s*c|
<=>(1+|s|)*|a| = (1+|s|)*|c|
<=> |a| = |c|
So,
o |a| + |b| = |c| + |d|
<=> |a| + |b| = |a| + |d|
<=> |b| = |d|

Bingo!... We are on the special case of:
o If |a| = |c|, then |b| = |d| that implies that A and B are in 4 possible points : the vertice of a square centered at 0(0,0)... (a,b), (-a,b), (a,-b) (-a,-b)

SUFF.


For some reason, I got stuck there. Thanks Fig. OA is C.
  [#permalink] 11 Dec 2007, 14:07
    Similar topics Author Replies Last post
Similar
Topics:
New posts EXPERTS_POSTS_IN_THIS_TOPIC In the rectangular coordinate system, are the points (a, b) jimjohn 2 17 Dec 2007, 14:57
New posts In the rectangular coordinate system, are the points (a, b) jimmyjamesdonkey 5 15 Jul 2008, 18:51
New posts In a rectangular coordinate system, are the points (a,b) and tarek99 8 15 Nov 2008, 15:56
Popular new posts 6 EXPERTS_POSTS_IN_THIS_TOPIC In the rectangular coordinate system, are the points (a, b) nsp007 14 10 Apr 2010, 23:25
This topic is locked, you cannot edit posts or make further replies. New EXPERTS_POSTS_IN_THIS_TOPIC In the rectangular coordinate system, are the points (a, b) mokap25 4 17 Aug 2011, 11:42
Display posts from previous: Sort by

In the rectangular coordinate system, are the points (a, b)

  Question banks Downloads My Bookmarks Reviews  


GMAT Club MBA Forum Home| About| Privacy Policy| 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®.