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

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Since PO and QO are perpendicular, t/s = - 1 / [#permalink] New post 15 Apr 2006, 14:55
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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
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Joined: 11 Nov 2005
Posts: 336
Location: London
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Kudos [?]: 8 [0], given: 0

 [#permalink] New post 15 Apr 2006, 15:36
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
Senior Manager
avatar
Joined: 08 Sep 2004
Posts: 258
Location: New York City, USA
Followers: 1

Kudos [?]: 9 [0], given: 0

 [#permalink] New post 15 Apr 2006, 18:33
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
  [#permalink] 15 Apr 2006, 18:33
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Since PO and QO are perpendicular, t/s = - 1 /

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