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In the figure above, points P

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ritula
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In the figure above, points P [#permalink] New post   13 Nov 2008, 03:35
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Re: In the figure above, points P [#permalink] New post   13 Nov 2008, 12:06
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prasun84 wrote:
good one really...
at 1st sight, s looks as \(sqrt3\)

radius=2
now, distance between P and Q=2\(sqrt2\)

now (s+\(sqrt3\))^2 + (t-1)^2 =8
to solve this equation i need \(sqrt3\) value for T, hence t=\(sqrt3\)

Hence s=1.
(if u look at the equation closely, it'll be clear....)


it s an interesting method we will have to rely on hit and trial to get the answer
My approach

Starting from the -ve X axix the order of angles formed will be 30,60,30,60
hence applying tan60 in the triangle created in the first quadrant will have sides hypotenuse= 2 base=1 (which is s) perpendicular =root 3(which is t)

hence s is 1
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Re: In the figure above, points P [#permalink] New post   13 Nov 2008, 08:01
good one really...
at 1st sight, s looks as \(sqrt3\)

radius=2
now, distance between P and Q=2\(sqrt2\)

now (s+\(sqrt3\))^2 + (t-1)^2 =8
to solve this equation i need \(sqrt3\) value for T, hence t=\(sqrt3\)

Hence s=1.
(if u look at the equation closely, it'll be clear....)
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Re: In the figure above, points P [#permalink] New post   13 Nov 2008, 20:00
calculate radius of the circle
radius^2= (sqrt(3))^2 + 1^2
radius = 2 (cant be -ve)

the equation of line passing through point P is,

y = -1/sqrt(3)x [ slope = (1-0)/(-sqrt(3) - 0) and y intercept of the line is 0)

The line passing through point Q is perpendicular to the above line
hence its equation is
y = sqrt(3)x
i.r t = sqrt(3)*s

Now OQ being radius can be expressed as
OQ^2 = s^2 + t^2
2^2 = s^2 + (sqrt(3)*s)^2
4 = S^2 * 4
S^2 = 1

S= 1 (being in Ist quadrant)
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Re: In the figure above, points P [#permalink] New post   13 Nov 2008, 21:42
Thats a very intelligent ans. but shldnt slope of OP be -1/sqrt(3)? (slope=difference in y/difference in x)
alpha_plus_gamma wrote:
calculate radius of the circle
radius^2= (sqrt(3))^2 + 1^2
radius = 2 (cant be -ve)

the equation of line passing through point P is,

y = -sqrt(3)x [ slope = (-sqrt(3) - 0) / (1-0) and y intercept of the line is 0)
The line passing through point Q is perpendicular to the above line
hence its equation is
y = sqrt(3)x
i.r t = sqrt(3)*s

Now OQ being radius can be expressed as
OQ^2 = s^2 + t^2
2^2 = s^2 + (sqrt(3)*s)^2
4 = S^2 * 4
S^2 = 1

S= 1 (being in Ist quadrant)
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Re: In the figure above, points P [#permalink] New post   13 Nov 2008, 21:48
Wow ! wht a shortcut. Applying angle formula never occured to me.
+1 from me
hibloom wrote:
prasun84 wrote:
good one really...
at 1st sight, s looks as \(sqrt3\)

radius=2
now, distance between P and Q=2\(sqrt2\)

now (s+\(sqrt3\))^2 + (t-1)^2 =8
to solve this equation i need \(sqrt3\) value for T, hence t=\(sqrt3\)

Hence s=1.
(if u look at the equation closely, it'll be clear....)


it s an interesting method we will have to rely on hit and trial to get the answer
My approach

Starting from the -ve X axix the order of angles formed will be 30,60,30,60
hence applying tan60 in the triangle created in the first quadrant will have sides hypotenuse= 2 base=1 (which is s) perpendicular =root 3(which is t)

hence s is 1
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alpha_plus_gamma
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Re: In the figure above, points P [#permalink] New post   13 Nov 2008, 21:55
ritula wrote:
Thats a very intelligent ans. but shldnt slope of OP be -1/sqrt(3)? (slope=difference in y/difference in x)


Yes you are correct. I have edited my post. thanks
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Re: In the figure above, points P [#permalink] New post   14 Nov 2008, 12:28
There's a much faster way to do this problem. To find the coordinates of q, all you have to do is switch the x- and y- values of the coordinates of P, then multiply the new y-value (t) by -1.

To understand why, imagine the full circle. Draw any two perpendicular lines through the center. They will intersect the circle at (a, b), (b, -a), (-a, -b), and (-b, a).
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Re: In the figure above, points P [#permalink] New post   14 Nov 2008, 12:45
Please check

7-t71146



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Re: In the figure above, points P [#permalink]
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