What set of transformations are applied to parallelogram ABCD to create A″B″C″D″?

Parallelogram formed by ordered pairs A at negative 4, 1, B at negative 3, 2, C at negative 1, 2, D at negative 2, 1. Second parallelogram transformed formed by ordered pairs A double prime at 4, negative 1, B double prime at 3, negative 2, C double prime at 1, negative 2, D double prime at 2, negative 1.
Reflected over the x-axis and reflected over the y-axis
Reflected over the y-axis and rotated 180°
Reflected over the x-axis and rotated 90° counterclockwise
Reflected over the y-axis and rotated 90° counterclockwise

1 answer

To determine the transformations that create parallelogram A″B″C″D″ from parallelogram ABCD, we start by analyzing the coordinates of the points.

Original Coordinates:

\( A(-4, 1) \)
\( B(-3, 2) \)
\( C(-1, 2) \)
\( D(-2, 1) \)

Transformed Coordinates:

\( A''(4, -1) \)
\( B''(3, -2) \)
\( C''(1, -2) \)
\( D''(2, -1) \)

Let's examine the result of the transformations step-by-step.

Reflection over the x-axis:
- Reflecting a point \( (x, y) \) over the x-axis gives \( (x, -y) \).
- Applying this to all original points:
  - \( A(-4, 1) \rightarrow A'(-4, -1) \)
  - \( B(-3, 2) \rightarrow B'(-3, -2) \)
  - \( C(-1, 2) \rightarrow C'(-1, -2) \)
  - \( D(-2, 1) \rightarrow D'(-2, -1) \)
Reflection over the y-axis:
- Now reflecting each of these new points over the y-axis (which gives \( (-x, y) \)):
  - \( A'(-4, -1) \rightarrow A''(4, -1) \)
  - \( B'(-3, -2) \rightarrow B''(3, -2) \)
  - \( C'(-1, -2) \rightarrow C''(1, -2) \)
  - \( D'(-2, -1) \rightarrow D''(2, -1) \)

After applying both transformations (reflection over the x-axis followed by reflection over the y-axis), we get the coordinates of parallelogram A″B″C″D″:

\( A''(4, -1) \)
\( B''(3, -2) \)
\( C''(1, -2) \)
\( D''(2, -1) \)

Thus, the transformations applied to parallelogram ABCD to create A″B″C″D″ are: Reflected over the x-axis and reflected over the y-axis.