Describing Rotations worksheet with four examples of geometric figures rotated on coordinate planes.

Let’s solve each rotation problem step by step.

We are given coordinate planes with a preimage (original shape) and its image after rotation. We need to describe the rotation in words — like “90° clockwise around the origin” or “180° counterclockwise around point X”.

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Problem ②:

Preimage: Rectangle WXYZ
Image: Rectangle W’X’Y’Z’

Look at the coordinates:

- W is at (-4, -2) → W’ is at (-2, 4)
- X is at (-1, -2) → X’ is at (-2, 1)
- Y is at (-1, -3) → Y’ is at (-3, 1)
- Z is at (-4, -3) → Z’ is at (-3, 4)

Wait — let’s check one point carefully.

Actually, looking again:

Original rectangle WXYZ:
- W(-4, -2), X(-1, -2), Y(-1, -3), Z(-4, -3)

Rotated image W’X’Y’Z’:
- W’(-2, 4), X’(-2, 1), Y’(-3, 1), Z’(-3, 4)

Hmm — that doesn’t look like a standard rotation about the origin. Let’s test if it’s rotated about a different point.

But wait — maybe I misread the labels.

Looking again at the diagram (mentally):

In problem ②, the original rectangle is on the left side of the y-axis, below the x-axis. The image is above the x-axis, still on the left? Wait no — actually, from the description in the first example, we’re rotating around the origin unless told otherwise.

Wait — let’s try rotating point W(-4, -2) 90° counterclockwise around origin:

Rule for 90° CCW: (x, y) → (-y, x)

So W(-4, -2) → (2, -4) — but that’s not where W’ is.

Try 90° clockwise: (x, y) → (y, -x)

W(-4, -2) → (-2, 4) — YES! That matches W’(-2, 4)

Check X(-1, -2) → (-2, 1) — yes, matches X’(-2, 1)

Y(-1, -3) → (-3, 1) — matches Y’(-3, 1)

Z(-4, -3) → (-3, 4) — matches Z’(-3, 4)

Perfect!

✔ So Problem ②: Rectangle WXYZ was rotated 90° clockwise around the origin to form rectangle W’X’Y’Z’.

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Problem ③:

Preimage: Rectangle DEFG
Image: Rectangle D’E’F’G’

Coordinates:

D(-7, -5), E(-3, -5), F(-3, -3), G(-7, -3)

Image:

D’(-1, 1), E’(3, 1), F’(3, 3), G’(-1, 3)

Wait — this looks like a translation? But the problem says “rotation”.

Wait — let’s check if it’s rotated 180° around some point.

Try rotating D(-7, -5) 180° around origin: becomes (7, 5) — not matching D’(-1,1)

Not origin.

Maybe around midpoint between D and D’?

Midpoint of D(-7,-5) and D’(-1,1): ((-7 + -1)/2, (-5+1)/2) = (-4, -2)

Test rotation 180° around (-4, -2):

Formula for 180° around (a,b): (x,y) → (2a - x, 2b - y)

So for D(-7,-5) around (-4,-2):

→ (2*(-4) - (-7), 2*(-2) - (-5)) = (-8 +7, -4 +5) = (-1, 1) → matches D’

Check E(-3,-5):

→ (2*(-4) - (-3), 2*(-2) - (-5)) = (-8+3, -4+5) = (-5,1) — but E’ is (3,1). Not matching.

Wait — maybe I have the wrong correspondence.

Perhaps the image is labeled differently? Let’s assume the order is preserved: D→D’, E→E’, etc.

Another idea: Maybe it’s rotated 180° around the point (-2, -1)? Let’s test.

Too messy.

Wait — look at the positions.

Original rectangle: from x=-7 to -3, y=-5 to -3 → bottom-left quadrant

Image: from x=-1 to 3, y=1 to 3 → top-right quadrant

This looks like a 180° rotation around the origin? Let’s test:

D(-7,-5) → (7,5) — not (-1,1)

No.

Wait — what if it’s rotated 180° around the point (-2, -1)?

Use formula: (x,y) → (2a - x, 2b - y) with a=-2, b=-1

D(-7,-5) → (2*(-2) - (-7), 2*(-1) - (-5)) = (-4 +7, -2 +5) = (3,3) — that’s F’, not D’

Not working.

Alternative approach: Look at vector from center.

Notice: From D(-7,-5) to D’(-1,1): change of (+6, +6)

From E(-3,-5) to E’(3,1): change of (+6, +6)

Same for others? F(-3,-3) to F’(3,3): (+6,+6)

G(-7,-3) to G’(-1,3): (+6,+6)

Oh! This is a TRANSLATION, not a rotation!

But the worksheet says “describing rotations”, and all problems should be rotations.

Wait — perhaps I misidentified the points.

Look back at the diagram description.

In problem ③, the preimage is rectangle DEFG, and image is D’E’F’G’.

But in the image, maybe the labeling is such that it's rotated 180° around a point.

Wait — another thought: What if it’s rotated 180° around the point (-2, -1)? Earlier calculation didn't work, but let's recalculate.

Set center C(h,k). For 180° rotation, the center is the midpoint of any point and its image.

Take D(-7,-5) and D’(-1,1): midpoint = ((-7-1)/2, (-5+1)/2) = (-4, -2)

Take E(-3,-5) and E’(3,1): midpoint = ((-3+3)/2, (-5+1)/2) = (0, -2) — not same as (-4,-2)

Contradiction.

Unless the correspondence is wrong.

What if D maps to G’? Or something else?

Perhaps the image is labeled in reverse order.

Assume that the rectangle is rotated 180°, so corners map to opposite corners.

Suppose D(-7,-5) maps to F’(3,3)? Midpoint: ((-7+3)/2, (-5+3)/2) = (-2, -1)

Check E(-3,-5) maps to G’(-1,3)? Midpoint: ((-3-1)/2, (-5+3)/2) = (-2, -1) — same!

F(-3,-3) maps to E’(3,1)? Midpoint: ((-3+3)/2, (-3+1)/2) = (0, -1) — not (-2,-1)

Not consistent.

F(-3,-3) to D’(-1,1): midpoint ((-3-1)/2, (-3+1)/2) = (-2, -1) — yes!

G(-7,-3) to E’(3,1): midpoint ((-7+3)/2, (-3+1)/2) = (-2, -1) — yes!

So if we pair:

D(-7,-5) ↔ F’(3,3)

E(-3,-5) ↔ G’(-1,3)

F(-3,-3) ↔ D’(-1,1)

G(-7,-3) ↔ E’(3,1)

All midpoints are (-2, -1)

So the rotation is 180° around the point (-2, -1)

And since 180° rotation is the same clockwise or counterclockwise, we can say either.

✔ So Problem : Rectangle DEFG was rotated 180° around the point (-2, -1) to form rectangle D’E’F’G’.

(Note: In many curricula, they might expect you to notice the center is the midpoint, and 180° rotation.)

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Problem ④:

Preimage: Parallelogram PQRS
Image: Parallelogram P’Q’R’S’

Coordinates:

P(1, -5), Q(4, -5), R(5, -2), S(2, -2)

Image:

P’(3, 3), Q’(6, 3), R’(7, 6), S’(4, 6)

Again, let’s see if it’s a translation: P to P’: (1→3, -5→3) = (+2, +8)

Q(4,-5) to Q’(6,3): (+2, +8) — same

R(5,-2) to R’(7,6): (+2, +8)

S(2,-2) to S’(4,6): (+2, +8)

Again, this is a translation, not a rotation!

But the worksheet says "rotations". There must be a mistake in my assumption.

Wait — perhaps the image is rotated, and I’m misreading the coordinates.

Maybe the parallelogram is rotated around a point.

Try assuming 90° rotation.

Suppose 90° clockwise around origin: (x,y) → (y, -x)

P(1,-5) → (-5, -1) — not (3,3)

No.

90° counterclockwise: (x,y) → (-y, x)

P(1,-5) → (5,1) — not (3,3)

180° around origin: (1,-5) → (-1,5) — not (3,3)

Now, try finding center of rotation.

Assume P(1,-5) maps to P’(3,3)

For a rotation, the center is equidistant, and angle is same.

The perpendicular bisector of PP’ should pass through center.

Midpoint of P and P’: ((1+3)/2, (-5+3)/2) = (2, -1)

Slope of PP’: (3 - (-5))/(3 - 1) = 8/2 = 4

Perpendicular slope: -1/4

So perpendicular bisector: line through (2,-1) with slope -1/4

Similarly, take Q(4,-5) to Q’(6,3)

Midpoint: (5, -1)

Slope QQ’: (3 - (-5))/(6-4) = 8/2 = 4

Perpendicular slope: -1/4

Perpendicular bisector: through (5,-1) with slope -1/4

These two lines are parallel? Both have slope -1/4, but different points — so they never meet? That can’t be.

Unless it’s not a rotation — but a translation.

But the worksheet insists on rotations.

Perhaps in problem ④, it’s rotated 180° around some point.

Let me calculate the vector from P to P’: (2,8)

If it were 180° rotation, the center would be midpoint, but as before, for different points, midpoints differ.

P to P’: midpoint (2,-1)

Q to Q’: midpoint (5,-1)

Different — so not 180° around a single point.

This suggests it’s not a rotation — but that can’t be right for the worksheet.

Wait — perhaps I have the wrong correspondence.

What if P maps to S’? Or something.

Another idea: Maybe the entire figure is rotated 90° around a point not the origin.

Let’s assume it’s 90° clockwise around some point (h,k).

The rule for 90° CW around (h,k):

(x,y) → (h + (y - k), k - (x - h)) = (h + y - k, k - x + h)

So for P(1,-5) → P’(3,3)

So:

3 = h + (-5) - k => h - k = 8 ...(1)

3 = k - 1 + h => h + k = 4 ...(2)

Add (1) and (2): 2h = 12 => h=6

Then from (2): 6 + k = 4 => k= -2

Check with Q(4,-5) → should go to (6 + (-5) - (-2), -2 - 4 + 6) = (6-5+2, -2-4+6) = (3,0) — but Q’ is (6,3). Not match.

Not working.

Try 90° CCW around (h,k):

Rule: (x,y) → (h - (y - k), k + (x - h)) = (h - y + k, k + x - h)

P(1,-5) → (h - (-5) + k, k + 1 - h) = (h+5+k, k+1-h) = (3,3)

So:

h + k + 5 = 3 => h + k = -2 ...(1)

k - h + 1 = 3 => -h + k = 2 ...(2)

Add (1) and (2): 2k = 0 => k=0

Then from (1): h + 0 = -2 => h= -2

Check Q(4,-5) → ( -2 - (-5) + 0, 0 + 4 - (-2) ) = (3, 6) — but Q’ is (6,3). Not match.

Still not.

Perhaps it's 180° around a point, and I need to find the correct pairing.

Assume that for a 180° rotation, the center is the midpoint of P and its image, but which image point corresponds to P?

Suppose P(1,-5) maps to R’(7,6)? Midpoint: ((1+7)/2, (-5+6)/2) = (4, 0.5)

Q(4,-5) maps to S’(4,6)? Midpoint: (4, 0.5) — same!

R(5,-2) maps to Q’(6,3)? Midpoint: ((5+6)/2, (-2+3)/2) = (5.5, 0.5) — not same.

R(5,-2) to P’(3,3): midpoint (4, 0.5) — yes!

S(2,-2) to Q’(6,3): midpoint (4, 0.5) — yes!

So if:

P(1,-5) ↔ R’(7,6)

Q(4,-5) ↔ S’(4,6)

R(5,-2) ↔ P’(3,3)

S(2,-2) ↔ Q’(6,3)

All midpoints are (4, 0.5)

So the rotation is 180° around the point (4, 0.5)

Since 0.5 is 1/2, we can write as (4, 1/2)

✔ So Problem ④: Parallelogram PQRS was rotated 180° around the point (4, 0.5) to form parallelogram P’Q’R’S’.

Or to avoid decimal, say (4, 1/2)

But in coordinate geometry, decimals are fine.

Some might prefer fractions.

But let's confirm with one point.

Rotate P(1,-5) 180° around (4, 0.5):

New x = 2*4 - 1 = 8 - 1 = 7

New y = 2*0.5 - (-5) = 1 + 5 = 6 → (7,6) which is R’ — good.

Q(4,-5) → 2*4 - 4 = 4, 2*0.5 - (-5) = 1 + 5 = 6 → (4,6) which is S’ — good.

Perfect.

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Now, summarize:

Problem ②: 90° clockwise around origin

Problem ③: 180° around (-2, -1)

Problem ④: 180° around (4, 0.5)

But in problem and ④, the centers are not nice integers, but mathematically correct.

Perhaps the worksheet expects us to recognize that for 180° rotation, the center is the midpoint, and report it.

Maybe in some cases, it's around the origin, but here it's not.

Another thought: In problem ③, if we consider the grid, the center (-2,-1) is a lattice point, and in ④, (4,0.5) is on a half-grid, which is acceptable.

So I think these are correct.

Final Answers:

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Final Answer:
② Rectangle WXYZ was rotated 90° clockwise around the origin to form rectangle W’X’Y’Z’.
③ Rectangle DEFG was rotated 180° around the point (-2, -1) to form rectangle D’E’F’G’.
④ Parallelogram PQRS was rotated 180° around the point (4, 0.5) to form parallelogram P’Q’R’S’.