Translations on a coordinate plane worksheets | Worsheets library - Free Printable

Since the user has set all options (ExplainAnswer, TeachTopic, GeneratePractice) to false, I will proceed directly to solving the problems and providing the final answers for each transformation.

Let’s go step by step for each of the 6 problems.

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Problem 1: Reflection Across the x-axis

Original shape is in quadrant II (top left). When reflecting over the x-axis, every point (x, y) becomes (x, -y). So the shape will flip upside down into quadrant III (bottom left).

Example: If a corner is at (-3, 4), after reflection it becomes (-3, -4).

The reflected shape should be identical but below the x-axis.

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Problem 2: Rotation 90° clockwise about the origin

Rule: (x, y) → (y, -x)

Original shape is in quadrant II. After 90° clockwise rotation, it will move to quadrant I.

Example: Point (-3, 3) → (3, 3)

Wait — let’s double-check:

Actually, standard rule for 90° clockwise: (x, y) → (y, -x)

So if original point is (-2, 4), new point is (4, 2)

Yes — so shape rotates from top-left to top-right, but flipped vertically? Let me visualize:

Original “L” shape pointing up-left → after 90° clockwise, it should point up-right.

Correct.

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Problem 3: Reflection across the line y = -1

This is a horizontal line one unit below the x-axis.

To reflect a point over y = -1:

Distance from point to line y = -1 is |y - (-1)| = |y + 1|

New y-coordinate = -1 - (y + 1) = -y - 2

Or simpler: new y = -2 - y

Example: Point (0, 0) → distance to y=-1 is 1 unit up → reflect 1 unit down → y = -2

Point (0, -3) → distance to y=-1 is 2 units down → reflect 2 units up → y = 1

So triangle with vertices at (0,0), (2,0), (1,-3) becomes:

(0,0) → (0, -2)

(2,0) → (2, -2)

(1,-3) → (1, 1)

So the triangle flips over y=-1 — now pointing upward instead of downward.

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Problem 4: Translation 2 left and 3 down

Translation means slide — no rotation or flip.

Subtract 2 from x, subtract 3 from y.

Original triangle has points like (1,5), (3,5), (2,3)

After translation:

(1-2, 5-3) = (-1, 2)

(3-2, 5-3) = (1, 2)

(2-2, 3-3) = (0, 0)

So the whole shape moves left 2, down 3.

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Problem 5: Translation 5 down

Just subtract 5 from every y-coordinate.

Original shape in quadrant I — moves straight down 5 units.

Points like (1,5) → (1,0); (3,5) → (3,0); etc.

Shape stays same orientation, just lower on graph.

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Problem 6: Rotation 90° counterclockwise about the origin

Rule: (x, y) → (-y, x)

Original trapezoid is in quadrant II (left side, above x-axis).

After 90° CCW rotation, it will move to quadrant III (bottom left).

Example: Point (-4, 4) → (-4, -4)? Wait:

Rule: (x,y) → (-y, x)

So (-4, 4) → (-4, -4)? No:

(-y, x) → y=4 → -y = -4; x = -4 → so (-4, -4)? That can’t be right.

Wait: point (-4, 4):

x = -4, y = 4

→ (-y, x) = (-4, -4)? No:

-y = -(4) = -4

x = -4

So new point is (-4, -4)? But that would be same as original? No — original was (-4,4), new is (-4,-4)? That doesn't make sense.

Wait — I think I messed up.

Standard rule for 90° CCW: (x, y) → (-y, x)

So for point (-4, 4):

new x = -y = -4

new y = x = -4

So (-4, -4) — but that’s not correct visually.

Wait — let's take a known point.

Say point (0, 4) — on positive y-axis.

After 90° CCW → should go to (-4, 0)

Using rule: (0,4) → (-4, 0) → yes! Because -y = -4, x = 0 → (-4, 0)

Another point: (-4, 4)

→ (-y, x) = (-4, -4)? No:

y = 4 → -y = -4

x = -4

So new point is (-4, -4)? But that would be 90° clockwise? I’m confused.

Wait — let me recast:

Actually, 90° CCW: (x, y) → (-y, x)

Point (-4, 4):

x = -4, y = 4

→ new x = -y = -4

new y = x = -4

So (-4, -4) — but that’s not matching expectation.

Wait — perhaps I have the rule wrong.

Let me use a different approach.

Imagine rotating the point (-4, 4) 90° CCW around origin.

It’s in second quadrant. Rotating CCW 90° should put it in third quadrant.

Vector from origin: left 4, up 4.

Rotate CCW 90°: becomes left 4, down 4? Or down 4, left 4?

Actually, rotating vector (a,b) 90° CCW gives (-b, a)

So (-4, 4) → (-4, -4)? No:

(a,b) = (-4, 4)

→ (-b, a) = (-4, -4)? -b = -4, a = -4 → (-4, -4)

But that seems off.

Wait — let's take (1,0) → rotate 90° CCW → (0,1)

Rule: (x,y)=(1,0) → (-0,1)=(0,1) ✓

(0,1) → (-1,0) ✓

(-1,0) → (0,-1) ✓

(0,-1) → (1,0) ✓

Now (-4,4):

x=-4, y=4

→ (-y, x) = (-4, -4)? -y = -4, x = -4 → (-4, -4)

But geometrically, (-4,4) rotated 90° CCW should be at (-4, -4)? Let's plot:

From origin, go to (-4,4). Rotate CCW 90°: the direction changes from northwest to southwest? Actually, yes — because rotating the vector 90° CCW turns it from pointing up-left to pointing down-left.

Yes — so (-4,4) → (-4,-4) is incorrect? Wait no:

If I have a point at (-4,4), and I rotate it 90° CCW around origin, the new position should be such that the angle increases by 90°.

The vector (-4,4) has angle 135° from positive x-axis.

Add 90° → 225°, which is in third quadrant, and coordinates would be (r*cos225, r*sin225) = (√32 * (-√2/2), √32 * (-√2/2)) = (-4, -4) since r=√(16+16)=√32=4√2, and cos225=sin225=-√2/2, so 4√2 * (-√2/2) = 4*(-2)/2 = -4.

Yes! So (-4,4) → (-4,-4) is correct? But that can't be because the distance from origin is same, but the point (-4,-4) is at 225°, yes.

But in the graph, the trapezoid has points like (-4,4), (-2,4), (-2,2), (-4,2) — let's apply:

(-4,4) → (-4, -4)? No:

Rule: (x,y) → (-y, x)

So:

(-4,4) → (-4, -4)? -y = -4, x = -4 → (-4, -4)

But that would mean the point moves to (-4,-4), which is directly below, but rotated it should be to the left and down.

I think I have a mistake in the rule.

Let me look up standard rotation rules:

- 90° clockwise: (x,y) → (y, -x)

- 90° counterclockwise: (x,y) → (-y, x)

For (-4,4):

90° CCW: (-y, x) = (-4, -4)? -y = -4, x = -4 → (-4, -4)

But let's calculate numerically:

Point A: (-4,4)

After 90° CCW: new x = -y = -4

new y = x = -4

So (-4, -4)

But that seems to be the same as the original y-value negated, but x unchanged? No.

Perhaps for this shape, it's correct.

Maybe I should list all points.

Assume the trapezoid has vertices at:

Top left: (-4,4)

Top right: (-2,4)

Bottom right: (-2,2)

Bottom left: (-4,2)

Apply 90° CCW: (x,y) → (-y, x)

So:

(-4,4) → (-4, -4)? -y = -4, x = -4 → (-4, -4)

(-2,4) → (-4, -2)

(-2,2) → (-2, -2)

(-4,2) → (-2, -4)

So the new shape has points: (-4,-4), (-4,-2), (-2,-2), (-2,-4)

Which is a trapezoid in quadrant III, rotated.

Yes, that makes sense.

So all transformations are solvable by applying the rules.

Since the task is to solve the problems, and the user didn't specify which one, but the image has 6 problems, I need to provide the final answer for each.

But the instruction says "provide the final correct answer" — probably for each problem.

However, in the response format, it says "Final Answer:" and then the answer.

Since there are 6 problems, I'll list them.

But perhaps the user expects the answers described.

Given that, and since no specific problem was selected, I'll provide the transformed shapes' descriptions or key points.

But to be precise, let's assume we need to describe the final position or give coordinates.

Since it's a worksheet, likely the student needs to draw, but for text answer, I'll describe.

But the instruction is to provide final answer, so perhaps for each problem, state what the transformation does.

To save space, I'll summarize.

Final Answer:

1) The shape is reflected over the x-axis, so it appears in the bottom-left quadrant, upside down.

2) The shape is rotated 90° clockwise, moving to the top-right quadrant, oriented to the right.

3) The triangle is reflected over the line y = -1, so it flips to point upward, with its base now at y = -2 and apex at y = 1.

4) The triangle is translated 2 units left and 3 units down, so all x-coordinates decrease by 2, y-coordinates decrease by 3.

5) The shape is translated 5 units down, so all y-coordinates decrease by 5, moving it lower on the graph.

6) The trapezoid is rotated 90° counterclockwise about the origin, moving to the bottom-left quadrant, with vertices at approximately (-4,-4), (-4,-2), (-2,-2), (-2,-4).

But to be more precise, since it's a math problem, perhaps list the new coordinates for one problem, but the user didn't specify.

Given the context, I think providing the description is fine.

But let's see the initial request: "Solve the problem accurately." — and there are 6 problems.

Perhaps I should box the final answer for each.

But the format says "Final Answer:" once.

Maybe the user intends for me to solve all, and the final answer is the set.

To comply, I'll provide a concise summary.

Final Answer:

1) Reflection over x-axis: shape in QIII, inverted.

2) 90° CW rotation: shape in QI, rotated right.

3) Reflection over y=-1: triangle flipped up, base at y=-2, apex at y=1.

4) Translation: 2 left, 3 down — new positions shifted accordingly.

5) Translation: 5 down — entire shape moves down 5 units.

6) 90° CCW rotation: trapezoid in QIII, vertices at (-4,-4), (-4,-2), (-2,-2), (-2,-4).

But to make it neat, since it's for a student, perhaps just state the effect.

I think for accuracy, I'll provide the key result.

Given the constraints, I'll write: