ktjorge wrote:
I think I understand how to solve the problem now, but I am having trouble understanding one assumption that people are making in many of the solutions: that O is at the origin. When I tried to solve this problem in the practice test I got to the point where I knew the radius was 2 and that line QO had a slope that was the negative reciprocal of line PO (since they were perpendicular bisectors), but I couldn't solve for slope m of line QO because I was stuck with 1=-sqrt(3)*m+b.
Can we assume that O is at the origin even though it isn't stated in the question, just because it looks like it is? Why can we assume that and not that the y-axis is equidistant from points P and Q (the diagram shows the y-axis passing directly in the middle of the 90 degree angle shown)?
O is the origin because it's on the intersection of x and y axis. Also, if we did not know that, we would not be able to solve the question and since it's a PS questions, we should be able to.
OFFICIAL GUIDE:Problem SolvingFigures: All figures accompanying problem solving questions are intended to provide information useful in solving the problems. Figures are drawn as accurately as possible. Exceptions will be clearly noted. Lines shown as straight are straight, and lines that appear jagged are also straight. The positions of points, angles, regions, etc., exist in the order shown, and angle measures are greater than zero. All figures lie in a plane unless otherwise indicated.
Data Sufficiency:Figures:• Figures conform to the information given in the question, but will not necessarily conform to the additional information given in statements (1) and (2).
• Lines shown as straight are straight, and lines that appear jagged are also straight.
• The positions of points, angles, regions, etc., exist in the order shown, and angle measures are greater than zero.
• All figures lie in a plane unless otherwise indicated.
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