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

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Senior Manager
Joined: 08 Sep 2004
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Since PO and QO are perpendicular, t/s = - 1 / [#permalink]

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15 Apr 2006, 15:55
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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

Kudos [?]: 25 [0], given: 0

Senior Manager
Joined: 11 Nov 2005
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Kudos [?]: 15 [0], given: 0

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15 Apr 2006, 16:36
B.

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

Kudos [?]: 15 [0], given: 0

Senior Manager
Joined: 08 Sep 2004
Posts: 257

Kudos [?]: 25 [0], given: 0

Location: New York City, USA

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

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

Kudos [?]: 25 [0], given: 0

15 Apr 2006, 19:33
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