Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 28 May 2017, 13:48

### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Since PO and QO are perpendicular, t/s = - 1 /

 post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
Senior Manager
Joined: 08 Sep 2004
Posts: 257
Location: New York City, USA
Followers: 1

Kudos [?]: 19 [0], given: 0

Since PO and QO are perpendicular, t/s = - 1 / [#permalink]

### Show Tags

15 Apr 2006, 15:55
00:00

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.

Since PO and QO are perpendicular,

t/s = - 1 / (1/-sqrt(3))

=> t/s = sqrt(3)

also t^2 + s^2 = 4

solving for s, we get 1.

B.

- Vipin
Senior Manager
Joined: 11 Nov 2005
Posts: 328
Location: London
Followers: 1

Kudos [?]: 14 [0], given: 0

### Show Tags

15 Apr 2006, 16:36
B.

OP=OQ, as radius = 2

P (-sqrt3,1) suggest that OP is at 60 degree from X axis (30 60 90 right angle traingle with 1, sqrt3, 2 sides)

therefore OQ is 30 degree from x axis. (again 30 60 90 right angle traingle, and we know that OQ =2, so Q must be 1, sqrt3)

vipin7um wrote:
Since PO and QO are perpendicular,

t/s = - 1 / (1/-sqrt(3)) how did you get this?
=> t/s = sqrt(3)

also t^2 + s^2 = 4

solving for s, we get 1.

B.

- Vipin
Senior Manager
Joined: 08 Sep 2004
Posts: 257
Location: New York City, USA
Followers: 1

Kudos [?]: 19 [0], given: 0

### Show Tags

15 Apr 2006, 19:33
SunShine wrote:
B.

OP=OQ, as radius = 2

P (-sqrt3,1) suggest that OP is at 60 degree from X axis (30 60 90 right angle traingle with 1, sqrt3, 2 sides)

therefore OQ is 30 degree from x axis. (again 30 60 90 right angle traingle, and we know that OQ =2, so Q must be 1, sqrt3)

vipin7um wrote:
Since PO and QO are perpendicular,

t/s = - 1 / (1/-sqrt(3)) how did you get this?
=> t/s = sqrt(3)

also t^2 + s^2 = 4

solving for s, we get 1.

B.

- Vipin

if two lines are perpendiculer, their slopes have the following relationship.

m1 / m2 = -1

so m1 = -1 / m2
15 Apr 2006, 19:33
Display posts from previous: Sort by

# Since PO and QO are perpendicular, t/s = - 1 /

 post reply Question banks Downloads My Bookmarks Reviews Important topics

 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®.