I'm stuck :( Can someone help me?

Question

Which sequence of transformations will map figure Q onto figure Q′?

Two congruent trapezoids figure Q and figure Q prime are drawn on a coordinate grid. Figure Q has vertices at (-9, 2), (-6, 4) (-4,4) and (-2,2). Figure Q prime has vertices at (2,2) (4,4) (6,4) and (9,2).

A. Translation of (x, y + 2), reflection over x = 1, and 180° rotation about the origin.

B. Translation of (x, y − 2), reflection over x = 1, and 180° rotation about the origin.

C. Translation of (x, y − 2), reflection over y = 1, and 180° rotation about the origin.

D. Translation of (x, y + 2), reflection over y = 1, and 180° rotation about the origin.

Answer 1

Let's work backward. Q' is in quadrant I and Q is in quadrant II.

180 degree rotation about the origin from Q' puts us in quadrant III with (-2,-2),(-4,-4),(-6,-4),(-9,-2)

Reflection across x = 1 puts us in quadrant IV. The given translations will not get us to quadrant II.

Reflection across y = 1 does put us in quadrant II, so our choices are C or D. We have (-2,4),(-4,6),(-6,6),(-9,4).

If we have translation of (x,y - 2), we get (-2,2),(-4,4),(-6,4),(-9,2). Since we were working backward, we'd do (x,y + 2) to Q.

The answer is D.

Answer 2

Hi Yariel V.,

From the coordinates given we can see that Q is in quadrant II and Q' is in quadrant I.

For A)

Q = (-9, 2), (-6, 4), (-4, 4), (-2, 2)

-Translation (x, y+2):

(-9, 4), (-6, 6), (-4, 6), (-2, 4)

-For reflection over x=1 (a vertical line at x=1), (-x+2, y):

(11, 4), (8, 6), (6, 6), (4, 4)

-Rotate 180⁰ (-x, -y):

(-11, -4), (-8, -6), (-6, -6), (-4, -4), this trapezoid Q is in quadrant III

For B)

Q = (-9, 2), (-6, 4), (-4, 4), (-2, 2)

-Translation (x, y-2):

(-9, 0), (-6, 2), (-4, 2), (-2, 0)

-For reflection over x=1, (-x+2, y):

(11, 0), (8, 2), (6, 2), (4, 0)

-Rotate 180⁰ (-x, -y):

(-11, 0), (-8, -2), (-6, -2), (-4, 0), this trapezoid Q is in quadrant III

For C)

Q = (-9, 2), (-6, 4), (-4, 4), (-2, 2)

-Translation (x, y-2):

(-9, 0), (-6, 2), (-4, 2), (-2, 0)

-For reflection over y=1, (x, -y+2):

(-9, 2), (-6, 0), (-4, 0), (-2, 2)

-Rotate 180⁰ (-x, -y):

(9, -2), (6, 0), (4, 0), (2, -2), this trapezoid Q is in quadrant IV

For D)

Q = (-9, 2), (-6, 4), (-4, 4), (-2, 2)

-Translation (x, y+2):

(-9, 4), (-6, 6), (-4, 6), (-2, 4)

-For reflection over x=1 (a vertical line at x=1), (x, -y+2):

(-9, -2), (-6, -4), (-4, -4), (-2, -2)

-Rotate 180⁰ (-x, -y):

(9, 2), (6, 4), (4, 4), (2, 2), this trapezoid Q is in quadrant I and is Q', our answer (D).

I hope this helps, Joe.