June 2022 Geometry Regents, Part 4

The following are some of the multiple questions from the recent January 2020 New York State Common Core Geometry Regents exam.

June 2022 Geometry, Part IV

A correct answer is worth up to 6 credits. Partial credit can be earned. Work must be shown or explained.

32. The coordinates of the vertices of D.ABC are A(-2,4), B(-7, -1), and C(-3, -3). Prove that D.ABC is isosceles.
[The use of the set of axes on the next page is optional.]

State the coordinates of D.A 'B 'C', the image of △ABC, after a translation 5 units to the right and 5 units down.

Prove that quadrilateral AA'C'C is a rhombus.
[The use of the set of axes below is optional.]

Answer:
The breakdown was 2 points from proving the triangle is isosceles, with one point for finding the lengths and one point for a statement saying it was isosceles and why. One point was given for finding all three set of coordinates of the image. The final three points were for proving that the object was a rhombus, including the work and a statement. If the student only showed that it was a parallelogram but not enough for a rhombus, that was one credit.

Points could be earned on the last portion proving that it was NOT a rhombus if incorrect coordinates for the image were found. In this case, supporting work still had to have been shown and a proper concluding statement needed to be given.

In the first part, you could use the distance formula, or you could use the graph and show it through the Pythagorean Theorem.

AB = √((5)² + (5)²) = √(50)
AC = √((1)² + (7)²) = √(50)

Triangle ABC is isosceles because AB ≅ AC.

To translate the pre-image to the image, add 5 to each x coordinate and subtract 5 from each y coordinate. You didn't have to show this, just the response.

A(-2,4) --> A'(3,-1)
B(-7,-1) --> B'(-2,-6)
C(-3,-3) --> C'(2, -8)

You didn't need to show your work on the graph. However, the visual is a great aid for solving the problem.

To prove that the shape was a rhombus, you had to show that four sides were congruent, OR you could show that the diagonals bisected each other. If you showed that the diagonals bisected each other, that only showed that it was a parallelogram. In that case, you still needed to show two consecutive sides were congruent.

Conversely, if you only showed two consecutive sides, then you still needed to show that it was a parallelogram.

The four sides of the rhomus are AA', A'C', C'C, and CA.

Length of AC = √(50). Shown above, you don't need to do it again..

Length of A'C' = √(50) because a translation is a rigid motion that preserves distance. This must be stated, or you could use the distance formula again.

Length of AA' = √( (5)² + (5)² ) = √(50)

Length of C'C = √( (5)² + (5)² ) = √(50)

All four sides are congruent, therefore AA'C'C is a rhombus.

Alternatively,

The slope of AC' is (-8 - 4)/(2 - -2) = -12/4 = -3

The slope of A'C is (-1 - -3)/(3 - -3) = 2/6 = 1/3

The slopes of the two diagonals are inverse reciprocals, so the diagonals are perpendicular. Therefore, the quadrilateral is a rhombus.

Other methods were possible.

End of Part Exam

How did you do?

Questions, comments and corrections welcome.

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