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A rectangle PQRS has its diagonal QS along the X-axis. Point

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A rectangle PQRS has its diagonal QS along the X-axis. Point [#permalink] New post 16 May 2004, 14:53
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A rectangle PQRS has its diagonal QS along the X-axis. Point Q is (6,0) and point S is (-6, 0). What are the coordinates of point P.
1) Diagonal PR passes through the origin
2) The measure of angle PQS is 30 degrees.
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 [#permalink] New post 16 May 2004, 15:27
A rectangle PQRS has its diagonal QS along the X-axis. Point Q is (6,0) and point S is (-6, 0). What are the coordinates of point P.
1) Diagonal PR passes through the origin
2) The measure of angle PQS is 30 degrees

E
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 [#permalink] New post 16 May 2004, 19:40
E it is

P could be on either side of the X axis.
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Re: DS-113 [#permalink] New post 17 May 2004, 02:36
hallelujah1234 wrote:
A rectangle PQRS has its diagonal QS along the X-axis. Point Q is (6,0) and point S is (-6, 0). What are the coordinates of point P.
1) Diagonal PR passes through the origin
2) The measure of angle PQS is 30 degrees.


1 is insufficient. That's clear.

2 is sufficient (because 1 => 2 and we have anlge PQS).
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 [#permalink] New post 17 May 2004, 14:37
Let P be (x, y)

PQ and PS are pependicular.
Therefore, (x-6)(x+6)+y^2 = 0

<PQS = 30degrees; so, PS = 6
or (x-6)^2+y^2 = 36

From the above two equations, x-6 = -3
or x = 3, y = 3sqrt(3) or -3sqrt(3)

P can be one of these: <3, 3sqrt(3)> or <3, -3sqrt(3)>

E is the answer
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 [#permalink] New post 17 May 2004, 17:56
hallelujah1234 wrote:
Let P be (x, y)

PQ and PS are pependicular.
Therefore, (x-6)(x+6)+y^2 = 0

<PQS = 30degrees; so, PS = 6
or (x-6)^2+y^2 = 36

From the above two equations, x-6 = -3
or x = 3, y = 3sqrt(3) or -3sqrt(3)

P can be one of these: <3, 3sqrt(3)> or <3, -3sqrt(3)>

E is the answer


Can we not just take Anand's approach and conclude both of the options together are insuff?
  [#permalink] 17 May 2004, 17:56
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