Corodinate Geom PS

Option D : 3^1/2

Here is my Solution:

Since it is a semi-circle therefore the radius remains equal, making it a 45-90-45 Degree Right angled triangle, symmetric about the y-axis. Hence, Option - D.

Algebric Approach

OP = OQ
(1-0)^2 + (-sqrt3-0)^2 = (s-0)^2 + (t-0)^2
1+3 = s^2+t^2

s^2+t^2 = 4 ---------(1)


Since the lines are perpendicular to each other, product of slopes = -1
[(1-0)/(-sqrt3-0)]*[(t-0)/(s-0)] = -1
t/s = sqrt3
t = s*sqrt3
From (1), s^2 + 3s^2 = 4
s^2 = 1
Since (s,t) is in first quadrant, s=1

Brent@GMATPrepNow wrote:

ani781 wrote:

Points P, Q lie on the circle with center O. What is the value of S ?
a) 1/2
b) 1
c) 2^(1/2)
d)3^(1/2)
e) (2^1/2)/2[/img]

Here's one approach:


So, s = 1
Answer: B

Cheers,
Brent

Hello Brent,

Thanks a lot for the excellent explanation.

Best Regards,
Sri

Brent@GMATPrepNow wrote:

ani781 wrote:

Points P, Q lie on the circle with center O. What is the value of S ?
a) 1/2
b) 1
c) 2^(1/2)
d)3^(1/2)
e) (2^1/2)/2[/img]

Here's one approach:


So, s = 1
Answer: B

Cheers,
Brent

Hello Brent,

Thanks a lot for the excellent explanation.

Best Regards,
Sri

Brent@GMATPrepNow wrote:

ani781 wrote:

Points P, Q lie on the circle with center O. What is the value of S ?
a) 1/2
b) 1
c) 2^(1/2)
d)3^(1/2)
e) (2^1/2)/2[/img]

Here's one approach:


So, s = 1
Answer: B

Cheers,
Brent

Hello Brent,

Thanks for the explanation. I was wondering that in the 3rd diagram that you have, once we know the value of OQ i.e. 2, what if we draw a straight line from P to Q and then use pythogoras theorem

i.e. 2^2 + 2^2 = PQ^2
=> PQ = 2 sq.root (2)

Hence, the distance from the Y axis to Q in line PQ is (2 sq. root (2))/2 = sq. root (2) and then this distance would the value of s. I was just wondering why this is in-correct? Thanks a lot for your help.

Best Regards,
Sri