Data Sufficiency: Graph-Geometry Probelm

[hijazim]

Any clue how to approach this problem?
The OA is B.
While I dont get it why A is not sufficient.

Any suggestions guys??

[kamu]

stmt 1 is useless because we already knew that the diagonals pass thru the origin.. the mid-point of (-6,0) (6.0)


stmt 2 gives a lot of details.

PQS = 30
therefore, PSQ = 60

since we know the angles.. on drawing lines at these angles we get the point of intersection, which is P.

I'm not sure if this is the correct explanation though.

[hijazim]

I've got it.

First before reviewing the statements 1 and 2, we can know that Diagonals of rectangle are equal and bisect each other
=> OP=OS=OQ=OR=QS/2=6

Statement 1: Gives us no new information as it is in the given QS passes through the origin and since O is the midpoint of QS then PR MUST pass thru O and O be its midpoint as diagonals of rectangle bisect each other.

Statement 2: Angle PQS=30 => Triangle SPQ is 30:60:90 triangle
=> SP=SQ/2=6

We already new before that OP=OS=6
Therefore, now we have OP=OS=SP=6
To simplify calculation and since we want to calculate coordinates of P we can say,
OP^2=SP^2 (x=x-coordinate of P, y=y-coordinate of P)
=> y^2 + x^2 = y^2 + (x+6)^2
=> x^2 = x^2+12x+36 (eliminate x^2)
=> 12x+36=0
=> x= -36/12 = -3 and y= 36 - x^2=36-9= 27 => y= 3sqrt(3)

Therefore, P [-3,3sqrt(3) ]

So, the answer is B.

Thanks for not helping guys. But I dont know if I face it in the exam i can eliminate A easily, but you need to know the approach to take to check if B is useful or go for D instead.

Thanks hope it was useful.

[kamu]

Calculating the coordinates is of no use.

[hijazim]

I know it is. But I calculated them just to prove they are calculable. But the problem is the hypothesis that you have to form. Anyway, it is a good question.