Now that I'm caught up with the current New York State Regents exams, I'm revisiting some older ones.
More Regents problems.
Geometry Regents, June 2011
Part III: Each correct answer will receive 4 credits. Partial credit is available.
35. On the set of coordinate axes below, graph the locus of points that are equidistant from the lines y = 6 and y = 2 and also graph the locus of points that are 3 units from the y-axis. State the
coordinates of all points that satisfy both conditions.
Answer:
The locus of points equidistant from horizontal lines y = 6 and y = 2 is the horizontal line y = 4, which is right in the middle of it. The locus of points 3 units from the y-axis is two vertical lines at x = 3 and x = -3.
The vertical lines intersect the horizontal line at (-3,4) and (3,4).
36. In the diagram below, tangent ML and secant MNK are drawn to circle O.
The ratio mLN : mNK : mKL is 3:4:5. Find m∠LMK.
Answer:
If the circle is divided into three arcs, then those three arcs must add up to 360 degree. If they are in the ratio of 3:4:5, then the following is true;
3x + 4x + 5x = 360
12x = 360
x = 30
This means that mLN = 3(30) = 90, mNK = 4(30) = 120, and mKL = 5(30) = 150.
The measure of angle LMK is (mKL - mLN) / 2 = (150 - 90) / 2 = 60 / 2 = 30 degrees.
2x2 − 4x = y + 1
x + y = 1
37. Solve the following system of equations graphically.
Answer:
Graph and label both lines. Label the points of intersection, and state the coordinates.
Rewrite the quadratic equation as y = 2x2 - 4x - 1, and the linear equation as y = -x + 1.
The Axis of Symmetry of the parabola is x = -(-4)/(2(2)) = 1. When x = 1, y = 2(1)2 - 4(1) - 1 = -3, so the vertex is (1,-3). You can start plotting points from there.
Your graph should look like the one below:
You may want to check on your calculator that 2(-1/2)2 - 4(-1/2) = (3/2) + 1 because you can't really go by "looks".
2(-1/2)2 - 4(-1/2) = 5/2
3/2 + 1 = 5/2
End of Part III.
More to come. Comments and questions welcome.
More Regents problems.
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