Compositions Answers

Triangle A B C is reflected across the line of reflection m to form triangle A prime B prime C prime. Triangle A prime B prime C prime is rotated about point B prime 270 degrees to form triangle A double-prime B double-prime C double-prime.

On a coordinate plane, 3 triangles are shown. Triangle B C D has points (1, 4), (1, 2), (5, 3). Triangle B prime C prime D prime has points (negative 1, 4), (negative 1, 2), (negative 5, 3). Triangle B double-prime C double-prime D double-prime has points (5, negative 1), (5, negative 3), (1, negative 2).

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Square A"B"C"D" is the final image after the rule was applied to square ABCD.

Parallelogram F"G"H"J" is the final image after the rule was applied to parallelogram FGHJ.

The rule is applied to ΔABC.

Trapezoid G H J K is rotated about G 90 degrees counterclockwise to form trapezoid G prime H prime J prime K prime. Trapezoid G prime H prime J prime K prime is reflected across the line of reflection m to form trapezoid G double-prime H double-prime J double-prime K double-prime.

Trapezoid E F G H is reflected across line of reflection k to form trapezoid E prime F prime G prime H prime. Trapezoid E prime F prime G prime H prime is shifted down to E double-prime F double-prime G double-prime H double-prime.

Which ordered pairs name the coordinates of vertices of the pre-image, trapezoid ABCD? Select two options.

On a coordinate plane, 3 parallelograms are shown. Parallelogram A B C D has points (3, 5), (6, 5), (4, 1), (1, 1). Parallelogram A prime B prime C prime D prime has points (3, negative 5), (6, negative 5), (4, negative 1), (1, negative 1). Parallelogram A double-prime B double-prime C double-prime D double-prime has points (negative 3, negative 4), (0, negative 4), (negative 2, 0), (negative 4, 0).