Geometry Problems of the Day (Geometry Regents, August 2012)

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, August 2012

Part III: Each correct answer will receive 4 credits. Partial credit is available.


35. As shown in the diagram below, the diagonals of parallelogram QRST intersect at E. If QE = x² + 6x, SE = x + 14, and TE = 6x - 1, determine TE algebraically.

Answer:

Since the diagonals of the parallelogram bisect each other, we know that QE = ES. Set up and solve this quadratic equation. Then use the value of x to find the length of TE.

x² + 6x = x + 14
x² + 5x - 14 = 0
(x + 7)(x - 2) = 0
x + 7 = 0 or x - 2 = 0
x = -7 or x = 2
Discard x = -7

Note that x = -7 is a valid answer for both QE and ES but creates an invalid answer for TE.

TE = 6(2) - 1 = 11.

36. The vertices of triangle RST are R(-6,5), S(-7,-2), and T(1,4).
The image of triangle RST after the composition T_–2,3 ° r_{y = x} is triangle R"S"T".
State the coordinates of triangle R"S"T".
[The use of the set of axes below is optional.]

Answer:

Remember with compositions, you do the second transformation first. You want to show the Translation OF THE Reflection, not a Translation followed by a Reflection.

To refect over the line y = x, switch the x-coordinate with the y-coordinate.

R(-6,5) --> R'(5, -6)
S(-7,-2) --> S'(-2,-7)
T(1,4) --> T'(4,1)

A Translation of (-2,3) means subtract 2 from each x-coordinate and add 3 to every y-coordinate.

R'(5, -6) --> R"(3,-3)
S'(-2,-7) --> S"(-4,-4)
T'(4,1) --> T"(2,4)

If you used the grid, make sure you show both steps and Label every point and state the final coordinates. Make sure your final answer is stated clearly.

37. On the set of axes below, solve the following system of equations graphically and state the coordinates of all points in the solution.

(x + 3)² + (y - 2)² = 25
2y + 4 = -x