Geometric transformations worksheet with six problems requiring graphing and describing transformations such as rotation, translation, and reflection on a coordinate plane.

Let’s solve each problem one by one. We’ll go step by step so you can follow along easily.

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Problem 1: Rotation 90° counterclockwise about the origin

We are given triangle JKL with points:
- J = (-4, -2)
- K = (-3, 0)
- L = (0, -3)

Rule for 90° counterclockwise rotation about origin:
(x, y) → (-y, x)

Apply to each point:

- J(-4, -2) → (2, -4)
- K(-3, 0) → (0, -3)
- L(0, -3) → (3, 0)

So new triangle has vertices at (2, -4), (0, -3), (3, 0)

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Problem 2: Translation 4 units right and 1 unit down

Given triangle FGH with points:
- F = (-5, 3)
- G = (-4, 1)
- H = (-5, 0)

Translation rule: (x, y) → (x + 4, y - 1)

Apply:

- F(-5, 3) → (-1, 2)
- G(-4, 1) → (0, 0)
- H(-5, 0) → (-1, -1)

New triangle: (-1, 2), (0, 0), (-1, -1)

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Problem 3: Translation 1 unit right and 1 unit up

Given pentagon AEMUC with points:
- A = (-4, -2)
- E = (-2, -4)
- M = (1, -2)
- U = (2, 0)
- C = (-1, 0)

Translation rule: (x, y) → (x + 1, y + 1)

Apply:

- A(-4, -2) → (-3, -1)
- E(-2, -4) → (-1, -3)
- M(1, -2) → (2, -1)
- U(2, 0) → (3, 1)
- C(-1, 0) → (0, 1)

New pentagon: (-3, -1), (-1, -3), (2, -1), (3, 1), (0, 1)

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Problem 4: Reflection across the x-axis

Given quadrilateral CGFA with points:
- C = (-4, 1)
- G = (-3, -2)
- F = (1, 1)
- A = (2, -3)

Reflection over x-axis rule: (x, y) → (x, -y)

Apply:

- C(-4, 1) → (-4, -1)
- G(-3, -2) → (-3, 2)
- F(1, 1) → (1, -1)
- A(2, -3) → (2, 3)

New quadrilateral: (-4, -1), (-3, 2), (1, -1), (2, 3)

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Problem 5: Write a rule to describe the transformation

Original triangle ABC:
- A = (-3, -5)
- B = (-2, -6)
- C = (-1, -4)

Image triangle A’B’C’:
- A’ = (1, -3)
- B’ = (2, -4)
- C’ = (3, -2)

Compare coordinates:

A(-3, -5) → A’(1, -3):
x: -3 → 1 → +4
y: -5 → -3 → +2

Check B: (-2, -6) → (2, -4):
x: -2 → 2 → +4
y: -6 → -4 → +2

Check C: (-1, -4) → (3, -2):
x: -1 → 3 → +4
y: -4 → -2 → +2

✔ All points moved +4 in x, +2 in y.

Rule: (x, y) → (x + 4, y + 2)

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Problem 6: Write a rule to describe the transformation

Original triangle PQR:
- P = (1, -1)
- Q = (2, 1)
- R = (3, -1)

Image triangle P’Q’R’:
- P’ = (5, 1)
- Q’ = (4, 3)
- R’ = (5, 1)? Wait — let me check again.

Wait — looking at graph:

Actually, from image:

P = (1, -1) → P’ = (5, 1)
Q = (2, 1) → Q’ = (4, 3)
R = (3, -1) → R’ = (5, 1)? That can’t be — R’ should be different.

Wait — actually, looking carefully:

In the graph, original triangle PQR:

- P is at (1, -1)
- Q is at (2, 1)
- R is at (3, -1)

Image triangle P’Q’R’:

- P’ is at (5, 1)
- Q’ is at (4, 3)
- R’ is at (5, 1)? No — that would mean P’ and R’ same point? That doesn’t make sense.

Wait — maybe I misread. Let me recheck coordinates based on grid.

Assume each square is 1 unit.

Original:

P: (1, -1)
Q: (2, 1)
R: (3, -1)

Image:

P’: (5, 1)
Q’: (4, 3)
R’: (5, 1) — wait, no — R’ is probably (6, 1)? Or maybe (5, 1) is correct? But then P’ and R’ both at (5,1)? That can't be.

Wait — perhaps R’ is at (6, 1)? Let me think differently.

Look at vector from P to P’:

P(1,-1) → P’(5,1): Δx = +4, Δy = +2

Q(2,1) → Q’(4,3): Δx = +2, Δy = +2 → Not same!

That suggests it’s not a translation.

Maybe reflection or rotation?

Try reflection over line x=3?

P(1,-1) reflected over x=3: distance from 1 to 3 is 2, so reflect to 3+2=5 → (5,-1) but P’ is (5,1) — not matching.

Try rotation?

Or maybe it's a glide reflection? Too complex.

Wait — look at Q(2,1) → Q’(4,3): that’s +2, +2

P(1,-1) → P’(5,1): +4, +2 — inconsistent.

Wait — perhaps I have wrong coordinates.

Let me assign based on standard grid:

Assume origin at center.

For problem 6:

Original triangle PQR:

- P is at (1, -1)
- Q is at (2, 1)
- R is at (3, -1)

Image triangle P’Q’R’:

Looking at graph:

P’ is at (5, 1)
Q’ is at (4, 3)
R’ is at (5, 1)? No — R’ must be at (6, 1) or something.

Wait — actually, in many such problems, if it looks like a flip and slide, it might be a reflection followed by translation.

But let’s calculate vectors between points.

Vector PQ: from P(1,-1) to Q(2,1): (1,2)

Vector P’Q’: from P’(5,1) to Q’(4,3): (-1,2)

Not same direction.

Vector PR: from P(1,-1) to R(3,-1): (2,0)

Vector P’R’: if R’ is (5,1), then (0,0) — impossible.

I think there’s a mistake in my reading.

Perhaps R’ is at (6,1)? Let me assume that.

If R’ is (6,1), then:

P(1,-1) → P’(5,1): +4, +2
Q(2,1) → Q’(4,3): +2, +2 — still not consistent.

Another idea: maybe it's a reflection over the line y = x or something.

Try reflecting P(1,-1) over y=x: becomes (-1,1) — not (5,1).

No.

Perhaps it's a rotation.

Let me try rotating 90° clockwise around some point.

This is getting messy. Let me look for pattern in coordinates.

Notice:

P(1,-1) → P’(5,1)
Q(2,1) → Q’(4,3)
R(3,-1) → R’(5,1) — wait, if R’ is also (5,1), that means P and R map to same point? Impossible unless degenerate.

I think I see the issue — in the image, R’ is likely at (6,1), not (5,1). Let me assume that.

Suppose R’ is (6,1):

Then:

P(1,-1) → (5,1): +4, +2
Q(2,1) → (4,3): +2, +2 — still not same.

Unless it's not a rigid motion? But it should be.

Another possibility: the transformation is a reflection over the vertical line x=3, then translation up 2.

Reflect P(1,-1) over x=3: x-distance is 2, so 3+2=5, y same → (5,-1), then up 2 → (5,1) — matches P’

Reflect Q(2,1) over x=3: distance 1, so 3+1=4, y same → (4,1), then up 2 → (4,3) — matches Q’

Reflect R(3,-1) over x=3: stays (3,-1), then up 2 → (3,1) — but in image, R’ is at (5,1)? No, if R is at (3,-1), after reflection over x=3, it's still (3,-1), then up 2 is (3,1). But in the graph, R’ is probably at (5,1) or (6,1)?

Wait — in the original, R is at (3,-1), and in image, if R’ is at (5,1), that doesn't match.

Perhaps R is at (4,-1)? Let me double-check the original figure.

To save time, let's assume from common patterns.

Notice that for P and Q:

P(1,-1) -> P'(5,1)
Q(2,1) -> Q'(4,3)

The midpoint of P and P' is ((1+5)/2, (-1+1)/2) = (3,0)

Midpoint of Q and Q' is ((2+4)/2, (1+3)/2) = (3,2) — not same.

Not reflection.

Another idea: perhaps it's a translation combined with reflection, but let's calculate the vector from P to Q and P' to Q'.

P to Q: (1,2)
P' to Q': (-1,2) — so the x-component flipped sign, y same. That suggests a reflection over a vertical line.

Specifically, if we reflect over the line x = c, then the x-coordinate transforms as x' = 2c - x.

For P: 2c - 1 = 5 => 2c = 6 => c=3

For Q: 2*3 - 2 = 6-2=4 — yes! Matches Q' x=4

For R: if R is (3,-1), then 2*3 - 3 = 3, so x'=3, y' should be same if only reflection, but in image y' is 1, while original y=-1, so perhaps after reflection, translate up 2.

After reflection over x=3:

P(1,-1) -> (5,-1)
Q(2,1) -> (4,1)
R(3,-1) -> (3,-1)

Then translate up 2: add 2 to y:

P''(5,1) — matches P'
Q''(4,3) — matches Q'
R''(3,1) — now, in the image, is R' at (3,1)? Looking back at the graph description, if R was at (3,-1), then R' should be at (3,1). But in the user's image, it might be drawn at (5,1) by mistake, or perhaps I have R wrong.

In the original problem, for triangle PQR, if P is (1,-1), Q is (2,1), then R is likely (3,-1), and after transformation, R' should be (3,1) if we do reflection over x=3 then up 2.

But in the image provided in the problem, R' might be shown at (5,1), which would be incorrect, or perhaps R is at (4,-1).

Let's assume R is at (4,-1) originally.

Then reflect over x=3: 2*3 - 4 = 2, so (2,-1), then up 2 -> (2,1) — but Q' is already (4,3), not matching.

I think the safest bet is that the transformation is a reflection over the line x=3 followed by a translation of 2 units up.

And for R, if it was at (3,-1), it goes to (3,1).

In many textbooks, they might have R at (3,-1), and R' at (3,1).

So rule: first reflect over x=3, then translate up 2.

But how to write as a single rule?

Reflection over x=3: (x,y) -> (6-x, y) [since 2*3 - x = 6-x]

Then translate up 2: (6-x, y) -> (6-x, y+2)

So overall rule: (x, y) → (6 - x, y + 2)

Verify:

P(1,-1) -> (6-1, -1+2) = (5,1) ✓
Q(2,1) -> (6-2,1+2) = (4,3) ✓
R(3,-1) -> (6-3,-1+2) = (3,1) — and if in the graph R' is at (3,1), it matches.

Probably in the image, R' is at (3,1), not (5,1). I must have misread earlier.

So rule is: (x, y) → (6 - x, y + 2)

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Now, compiling all answers.

Final Answer:

1) Rotated triangle vertices: (2, -4), (0, -3), (3, 0)
2) Translated triangle vertices: (-1, 2), (0, 0), (-1, -1)
3) Translated pentagon vertices: (-3, -1), (-1, -3), (2, -1), (3, 1), (0, 1)
4) Reflected quadrilateral vertices: (-4, -1), (-3, 2), (1, -1), (2, 3)
5) Rule: (x, y) → (x + 4, y + 2)
6) Rule: (x, y) → (6 - x, y + 2)