Question

the graph shows parallelogram cdef. cdef is rotated 270° clockwise about the origin to produce cdef. select all of the statements that are true. d’ and e’ have the same x - coordinate. the distance between c’ and f’ is 3 units. the x - coordinate of c’ is negative. the x - and y - coordinates of d’ have different signs.

Explanation:

To solve this, we first find the coordinates of the original points:

• \( C(3, 3) \), \( D(4, 7) \), \( E(7, 7) \), \( F(6, 3) \)

The rule for a \( 270^\circ \) clockwise rotation about the origin is \( (x, y) \to (y, -x) \).

Step 1: Rotate each point

• For \( C(3, 3) \): \( C'(3, -3) \)
• For \( D(4, 7) \): \( D'(7, -4) \)
• For \( E(7, 7) \): \( E'(7, -7) \)
• For \( F(6, 3) \): \( F'(3, -6) \)

Step 2: Analyze each statement

1. \( D' \) and \( E' \) have the same \( x \)-coordinate.

\( D'(7, -4) \) and \( E'(7, -7) \) both have \( x = 7 \). True.

1. The distance between \( C' \) and \( F' \) is 3 units.

\( C'(3, -3) \), \( F'(3, -6) \). Since \( x \)-coordinates are equal, distance is \( |-3 - (-6)| = 3 \). True.

1. The \( x \)-coordinate of \( C' \) is negative.

\( C'(3, -3) \) has \( x = 3 \) (positive). False.

1. The \( x \)- and \( y \)-coordinates of \( D' \) have different signs.

\( D'(7, -4) \): \( x = 7 \) (positive), \( y = -4 \) (negative). Different signs. True.

Answer:

• \( D' \) and \( E' \) have the same \( x \)-coordinate. (True)
• The distance between \( C' \) and \( F' \) is 3 units. (True)
• The \( x \)- and \( y \)-coordinates of \( D' \) have different signs. (True)

(Note: The statement "The \( x \)-coordinate of \( C' \) is negative" is false.)