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

It is currently 15 Sep 2014, 01:26

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

ps-coordinate geometry

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
VP
VP
avatar
Joined: 30 Sep 2004
Posts: 1493
Location: Germany
Followers: 4

Kudos [?]: 49 [0], given: 0

GMAT Tests User
ps-coordinate geometry [#permalink] New post 31 May 2005, 01:25
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
What is the least possible distance between a point on the circle x^2 + y^2 = 1 and a point on the line y = 3/4*x - 3?

A) 1.4
B) sqrt (2)
C) 1.7
D) sqrt (3)
E) 2.0
_________________

If your mind can conceive it and your heart can believe it, have faith that you can achieve it.

Director
Director
avatar
Joined: 18 Feb 2005
Posts: 674
Followers: 1

Kudos [?]: 2 [0], given: 0

GMAT Tests User
 [#permalink] New post 31 May 2005, 03:47
x^2 + y^2 = 1 is a circle with centre at (0,0) and y = 3/4*x - 3 is a line which doesnt pass through origin . So make x = 0 we get y = -3 which is 2 units away from the circle which intercepts the X -axis at -1 . So the distance is 2.

Also check with y=0 we get x=4 which is 3units away from the circle.


So E is the answer
Current Student
avatar
Joined: 28 Dec 2004
Posts: 3405
Location: New York City
Schools: Wharton'11 HBS'12
Followers: 13

Kudos [?]: 159 [0], given: 2

GMAT Tests User
 [#permalink] New post 31 May 2005, 05:33
E it is...

just draw out the 2 and you will see....
VP
VP
avatar
Joined: 18 Nov 2004
Posts: 1447
Followers: 2

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

GMAT Tests User
 [#permalink] New post 31 May 2005, 05:56
"A"

draw a perpendicular from the center to the line....distance from the center to the line is: 3/sqrt(25/16) = 12/5

min distance = 12/5 - radius of circle = 12/5 -1 = 7/5 = 1.4
VP
VP
avatar
Joined: 30 Sep 2004
Posts: 1493
Location: Germany
Followers: 4

Kudos [?]: 49 [0], given: 0

GMAT Tests User
 [#permalink] New post 31 May 2005, 06:11
banerjeea_98 wrote:
"A"

draw a perpendicular from the center to the line....distance from the center to the line is: 3/sqrt(25/16) = 12/5

min distance = 12/5 - radius of circle = 12/5 -1 = 7/5 = 1.4


baner, it seems to me that you know for almost every problem the right formula. :-D do you have a list of formulas ?
_________________

If your mind can conceive it and your heart can believe it, have faith that you can achieve it.

VP
VP
avatar
Joined: 18 Nov 2004
Posts: 1447
Followers: 2

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

GMAT Tests User
 [#permalink] New post 31 May 2005, 06:46
christoph wrote:
banerjeea_98 wrote:
"A"

draw a perpendicular from the center to the line....distance from the center to the line is: 3/sqrt(25/16) = 12/5

min distance = 12/5 - radius of circle = 12/5 -1 = 7/5 = 1.4


baner, it seems to me that you know for almost every problem the right formula. :-D do you have a list of formulas ?


Altho I don't have a list, but I think in GMAT coordinate geometry, u need to know only three formulas.

1. Finding the slope of a line with two points (x1,y1) (x2,y2) =
(y2-y1)/(x2-x1)

2. Distance of a point (x1,y1) from a line (ax+by+c = 0) :
(ax1+by1+c)/sqrt(a^2+b^2)

3. Distance between two points = sqrt{(x1-x2)^2 + (y1-y2)^2}
Intern
Intern
avatar
Joined: 06 Apr 2004
Posts: 42
Followers: 0

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

 [#permalink] New post 31 May 2005, 11:12
banerjeea_98 wrote:
"A"

draw a perpendicular from the center to the line....distance from the center to the line is: 3/sqrt(25/16) = 12/5

min distance = 12/5 - radius of circle = 12/5 -1 = 7/5 = 1.4


bravo, baner... really good approach. Agree on A.
Senior Manager
Senior Manager
User avatar
Joined: 21 Mar 2004
Posts: 445
Location: Cary,NC
Followers: 2

Kudos [?]: 6 [0], given: 0

GMAT Tests User
Re: ps-coordinate geometry [#permalink] New post 31 May 2005, 13:45
christoph wrote:
What is the least possible distance between a point on the circle x^2 + y^2 = 1 and a point on the line y = 3/4*x - 3?

A) 1.4
B) sqrt (2)
C) 1.7
D) sqrt (3)
E) 2.0



Draw the circle and the triangle on paper.

The triangle is a 3-4-5 triangle. The height from right angle to hypotenuse = 3*4/5
=2.4

Final Ans is 2.4 -radius = 1.4

This formula might be helpful for right angle triangles.
Height = side1*side2 / hypotenuse
_________________

ash
________________________
I'm crossing the bridge.........

Director
Director
avatar
Joined: 18 Feb 2005
Posts: 674
Followers: 1

Kudos [?]: 2 [0], given: 0

GMAT Tests User
 [#permalink] New post 31 May 2005, 19:17
banerjeea_98 wrote:
"A"

draw a perpendicular from the center to the line....distance from the center to the line is: 3/sqrt(25/16) = 12/5

min distance = 12/5 - radius of circle = 12/5 -1 = 7/5 = 1.4


Good one Baner....Forgot the concept...Thanks for reminding
  [#permalink] 31 May 2005, 19:17
    Similar topics Author Replies Last post
Similar
Topics:
PS-coordinates lan583 5 12 Sep 2006, 04:14
Geometry ywilfred 5 16 Oct 2005, 07:50
Geometry Macedon 3 12 Oct 2005, 05:00
geometry mirhaque 13 21 Feb 2005, 15:56
Geometry crackgmat750 10 02 Oct 2004, 22:06
Display posts from previous: Sort by

ps-coordinate geometry

  Question banks Downloads My Bookmarks Reviews Important topics  


cron

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®.