Transformations on the Coordinate Plane Worksheet | 8th Grade PDF ... - Free Printable

Let’s solve each problem step by step. We’ll go one at a time, carefully tracking the coordinates of each shape after every transformation.

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Problem 1: Shape A → reflected in y-axis → translated by vector (-2, -6) → Shape B

Shape A has vertices (from grid):
- Top-left: (-5, 4)
- Top-right: (-3, 4)
- Bottom-right: (-3, 1)
- Bottom-left: (-4, 1) — wait, actually looking again: it's an L-shape.

Actually, let’s list all corner points of Shape A:

Looking at the grid:
- Left top: (-5, 4)
- Right top: (-3, 4)
- Right bottom of top part: (-3, 3)
- Then down to (-3, 1)
- Then left to (-4, 1)
- Then up to (-4, 3)? Wait — no, better to trace the outline.

Actually, from the image, Shape A is made of two rectangles:

Top rectangle: from x=-5 to x=-3, y=3 to y=4 → so corners: (-5,4), (-3,4), (-3,3), (-5,3)

Bottom rectangle: from x=-4 to x=-3, y=1 to y=3 → corners: (-4,3), (-3,3), (-3,1), (-4,1)

But since (-3,3) and (-4,3) are shared, overall vertices for drawing: we can use outer corners:

Key points to transform:
- (-5, 4)
- (-3, 4)
- (-3, 1)
- (-4, 1)
- (-4, 3) — but maybe not needed if we just track the main corners.

Actually, simplest: take the 4 outermost corners that define the shape:

Let’s pick:
P1 = (-5, 4)
P2 = (-3, 4)
P3 = (-3, 1)
P4 = (-4, 1)

Wait — from (-3,1) to (-4,1) is bottom, then up to (-4,3), then to (-5,3), then to (-5,4). So actually 6 points? But for reflection and translation, we can transform key points.

To avoid confusion, let’s define the shape by its bounding box or just transform each vertex.

Actually, let’s list all unique vertices of Shape A:

From the grid:
- (-5, 4)
- (-3, 4)
- (-3, 3)
- (-3, 1)
- (-4, 1)
- (-4, 3)
- (-5, 3)

But many are redundant. For transformation, we only need to transform the “corner” points that will define the new shape.

Easier: reflect each point over y-axis first.

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

So:

Original points of Shape A (let’s take the 4 extreme points that form the L):

Actually, let’s take these 4 points that define the shape clearly:

A1: (-5, 4)
A2: (-3, 4)
A3: (-3, 1)
A4: (-4, 1)

Wait — from (-3,1) to (-4,1) is correct, but then from (-4,1) up to (-4,3), then to (-5,3), then to (-5,4). So perhaps we should include (-4,3) and (-5,3).

But to keep it simple, let’s transform all distinct vertices:

List of vertices for Shape A (going clockwise from top-left):

1. (-5, 4)
2. (-3, 4)
3. (-3, 3)
4. (-3, 1)
5. (-4, 1)
6. (-4, 3)
7. (-5, 3) — back to start? Actually, from (-4,3) to (-5,3) to (-5,4).

So 7 points? That’s messy.

Notice: the shape is symmetric in a way, but let’s do this:

After reflection over y-axis, each point (x,y) becomes (-x,y).

So:

(-5,4) → (5,4)
(-3,4) → (3,4)
(-3,3) → (3,3)
(-3,1) → (3,1)
(-4,1) → (4,1)
(-4,3) → (4,3)
(-5,3) → (5,3)

Now, translate by vector (-2, -6): add -2 to x, -6 to y.

So:

(5,4) → (3, -2)
(3,4) → (1, -2)
(3,3) → (1, -3)
(3,1) → (1, -5)
(4,1) → (2, -5)
(4,3) → (2, -3)
(5,3) → (3, -3)

Now, plot these points to get Shape B.

The transformed shape will have:

- From (3,-2) to (1,-2) to (1,-3) to (1,-5) to (2,-5) to (2,-3) to (3,-3) and back to (3,-2)? Let’s see the connections.

Actually, the original shape had:

From (-5,4) to (-3,4) to (-3,3) to (-3,1) to (-4,1) to (-4,3) to (-5,3) to (-5,4)

After reflection and translation, the order is preserved:

Start at (3,-2) [was (-5,4)]
→ (1,-2) [was (-3,4)]
→ (1,-3) [was (-3,3)]
→ (1,-5) [was (-3,1)]
→ (2,-5) [was (-4,1)]
→ (2,-3) [was (-4,3)]
→ (3,-3) [was (-5,3)]
→ back to (3,-2) [was (-5,4)]

So Shape B is drawn with those vertices.

We don’t need to draw here, but for answer, we can describe or just state the final positions. Since the question says "complete the transformations by drawing", but as text, we’ll describe the final shape’s position.

But for the purpose of this exercise, since we’re to provide final answer, and it’s multiple problems, perhaps we list the final coordinates or describe.

But the instruction is to solve accurately, and since it’s a drawing task, maybe we just confirm the steps.

However, for the Final Answer section, we need to box something. Perhaps for each problem, we can state the key result.

But let’s continue with all problems similarly.

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Problem 2: Triangle C rotated 180° about (-1,2), then reflected in y=x to give Triangle D

First, find vertices of Triangle C.

From grid: it’s a right triangle with vertices at:

- (0,3) — top
- (0,0) — origin? Wait, no.

Looking: base on x-axis from (0,0) to (4,0), and height to (0,3)? No.

Actually, from image: Triangle C has vertices at:

- (0,3) — top left
- (0,0) — bottom left? But (0,0) is origin, and it goes to (4,0)? Let’s see.

The triangle is shaded: from (0,3) down to (0,0)? No, the line is from (0,3) to (4,0), and also to (0,0)? Actually, it’s a triangle with vertices at (0,3), (0,0), and (4,0)? But (0,0) to (4,0) is base, (0,0) to (0,3) is height, but then hypotenuse from (0,3) to (4,0).

In the grid, it’s shown with points at (0,3), (4,0), and (0,0)? But (0,0) is included? Let me check.

Actually, looking closely: the triangle has vertices at:

- (0,3)
- (4,0)
- and (0,0)? But the side from (0,3) to (4,0) is the hypotenuse, and legs along axes? From (0,3) to (0,0) is vertical, (0,0) to (4,0) is horizontal, so yes, vertices: A(0,3), B(0,0), C(4,0)

But in the label, it’s called Triangle C, and it’s in the first quadrant.

Confirm: from (0,3) to (0,0) to (4,0) back to (0,3)? Yes.

So vertices: P1(0,3), P2(0,0), P3(4,0)

Now, rotate 180° about point (-1,2).

Rotation 180° about a point (a,b): the formula is:

(x,y) → (2a - x, 2b - y)

Because 180° rotation is same as reflection through the point.

So center (a,b) = (-1,2)

For each point:

P1(0,3) → (2*(-1) - 0, 2*2 - 3) = (-2 -0, 4-3) = (-2,1)

P2(0,0) → (2*(-1) - 0, 2*2 - 0) = (-2, 4)

P3(4,0) → (2*(-1) - 4, 2*2 - 0) = (-2-4, 4-0) = (-6,4)

So after rotation, vertices are: (-2,1), (-2,4), (-6,4)

Now, reflect this triangle in the line y = x.

Reflection over y=x swaps x and y coordinates.

So:

(-2,1) → (1,-2)

(-2,4) → (4,-2)

(-6,4) → (4,-6)

So Triangle D has vertices at (1,-2), (4,-2), (4,-6)

We can verify: it should be a right triangle now in fourth quadrant.

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Problem 3: Shape E enlarged by scale factor -2 from center (2,-3), then translated by vector (-4,3) to give Shape F

First, find vertices of Shape E.

From grid: it’s a T-shape or something? Looking:

Shape E is at bottom: from x=1 to x=3, y=-5 to y=-4? And a stem up.

Specifically:

- Base: from (1,-5) to (3,-5) to (3,-4) to (1,-4)? No.

Actually, it’s like a plus but missing top? Let’s see:

Points:
- (1,-5), (3,-5), (3,-4), (2,-4), (2,-3), (1,-3)? Not clear.

From image: Shape E has:

- Bottom rectangle: x=1 to 3, y=-5 to -4 → so corners (1,-5),(3,-5),(3,-4),(1,-4)
- Top part: from x=2 to 2? Wait, it’s a single column up: from (2,-4) to (2,-3)

So vertices: let’s list:

A(1,-5), B(3,-5), C(3,-4), D(2,-4), E(2,-3), F(1,-3), and back? From (1,-3) to (1,-4)? But (1,-4) is already there.

Actually, the shape is: starting from (1,-5) to (3,-5) to (3,-4) to (2,-4) to (2,-3) to (1,-3) to (1,-4) to (1,-5)? That would be self-intersecting.

Better: it’s a cross without top arm? Standard T-shape upside down.

Typically: base from x=1 to 3 at y=-5, and stem from x=2, y=-4 to y=-3.

So the boundary:

- Start at (1,-5)
- to (3,-5)
- to (3,-4)
- to (2,-4) [since stem starts]
- to (2,-3)
- to (1,-3) ? But then how to close? From (1,-3) to (1,-4) to (1,-5)? But (1,-4) is not directly connected.

Perhaps the vertices are: (1,-5), (3,-5), (3,-4), (2,-4), (2,-3), (1,-3), and then back to (1,-5) via (1,-4), but that’s not efficient.

For transformation, we can take key points: the corners.

Let’s take these points for Shape E:

P1(1,-5) // bottom left
P2(3,-5) // bottom right
P3(3,-4) // right middle
P4(2,-4) // stem right
P5(2,-3) // top of stem
P6(1,-3) // left top? But then from (1,-3) to (1,-4) to (1,-5)

Actually, to avoid missing, let’s include (1,-4) as well.

But notice that from (1,-3) to (1,-4) is vertical, and (1,-4) to (1,-5) is another segment.

So full set: (1,-5), (3,-5), (3,-4), (2,-4), (2,-3), (1,-3), (1,-4)

And back to (1,-5).

Now, enlarge by scale factor -2 from center (2,-3).

Scale factor negative means enlargement and inversion through the center.

Formula for enlargement with scale factor k from center (a,b):

New point = (a + k*(x-a), b + k*(y-b))

Here k = -2, center (2,-3)

So for each point (x,y):

New x = 2 + (-2)*(x - 2) = 2 -2(x-2) = 2 -2x +4 = 6 -2x

New y = -3 + (-2)*(y - (-3)) = -3 -2(y+3) = -3 -2y -6 = -9 -2y

Let’s compute for each vertex:

P1(1,-5):
x' = 6 -2*1 = 4
y' = -9 -2*(-5) = -9 +10 = 1
→ (4,1)

P2(3,-5):
x' = 6 -2*3 = 6-6=0
y' = -9 -2*(-5) = -9+10=1
→ (0,1)

P3(3,-4):
x' = 6 -6=0
y' = -9 -2*(-4) = -9+8= -1
→ (0,-1)

P4(2,-4):
x' = 6 -4=2? 6 -2*2=6-4=2
y' = -9 -2*(-4)= -9+8= -1
→ (2,-1)

P5(2,-3):
x' = 6 -4=2
y' = -9 -2*(-3)= -9+6= -3
→ (2,-3) // center stays? Scale factor -2, but center is fixed? No, in enlargement, the center is fixed only if scale factor is defined from it, but here the point (2,-3) is the center, so it should map to itself? Let's see:

x' = 6 -2*2 = 6-4=2
y' = -9 -2*(-3) = -9+6= -3 → yes, (2,-3) maps to itself.

P6(1,-3):
x' = 6 -2*1=4
y' = -9 -2*(-3)= -9+6= -3
→ (4,-3)

P7(1,-4):
x' = 6 -2*1=4
y' = -9 -2*(-4)= -9+8= -1
→ (4,-1)

So after enlargement, vertices are:
(4,1), (0,1), (0,-1), (2,-1), (2,-3), (4,-3), (4,-1)

Now, translate by vector (-4,3): add -4 to x, +3 to y.

So:

(4,1) → (0,4)
(0,1) → (-4,4)
(0,-1) → (-4,2)
(2,-1) → (-2,2)
(2,-3) → (-2,0)
(4,-3) → (0,0)
(4,-1) → (0,2)

So Shape F has these vertices.

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Problem 4: Shape G rotated 90° clockwise about origin, then reflected in x=-3 to give Shape H

First, vertices of Shape G.

From grid: it’s a trapezoid or quadrilateral.

Points:
- (-4,-3) — top left?
- (-2,-3) — top right?
- (-1,-5) — bottom right?
- (-4,-5) — bottom left?

Looking: from (-4,-3) to (-2,-3) to (-1,-5) to (-4,-5) back to (-4,-3)? But (-4,-5) to (-4,-3) is vertical.

Actually, it might be: (-4,-3), (-2,-3), (-1,-5), (-4,-5)

Yes, that makes a quadrilateral.

So vertices: A(-4,-3), B(-2,-3), C(-1,-5), D(-4,-5)

Rotate 90° clockwise about origin (0,0).

Rule for 90° clockwise: (x,y) → (y, -x)

So:

A(-4,-3) → (-3, 4) [because y=-3, -x=4]
B(-2,-3) → (-3, 2)
C(-1,-5) → (-5, 1)
D(-4,-5) → (-5, 4)

So after rotation: (-3,4), (-3,2), (-5,1), (-5,4)

Now, reflect in the line x = -3.

Reflection over vertical line x = a: the formula is x' = 2a - x, y' = y

Here a = -3, so x' = 2*(-3) - x = -6 - x, y' = y

So for each point:

(-3,4) → x' = -6 - (-3) = -3, y'=4 → (-3,4) // on the line, stays
(-3,2) → (-3,2) // also on line? x=-3, so yes
(-5,1) → x' = -6 - (-5) = -1, y'=1 → (-1,1)
(-5,4) → x' = -6 - (-5) = -1, y'=4 → (-1,4)

So Shape H has vertices: (-3,4), (-3,2), (-1,1), (-1,4)

Note: the first two points are on the line, so unchanged; the other two are reflected.

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Problem 5: Triangle I enlarged by scale factor 1/2, center (0,-1), then reflected in y=-x to give Triangle J

First, vertices of Triangle I.

From grid: it’s a triangle with vertices at:

- (-2,5) — top left?
- (4,5) — top right?
- (4,1) — bottom right?

Looking: from (-2,5) to (4,5) to (4,1) back to (-2,5)? But that would be a right triangle? Distance from (-2,5) to (4,1) is diagonal.

Actually, in the grid, it’s shown with points at (-2,5), (4,5), and (4,1)? But then the third side is from (4,1) to (-2,5).

Yes, so vertices: A(-2,5), B(4,5), C(4,1)

Enlarge by scale factor 1/2 from center (0,-1).

Formula: new point = (a + k*(x-a), b + k*(y-b)) with k=1/2, a=0, b=-1

So:

x' = 0 + (1/2)(x - 0) = x/2

y' = -1 + (1/2)(y - (-1)) = -1 + (1/2)(y+1)

Compute:

A(-2,5):
x' = -2/2 = -1
y' = -1 + (1/2)(5+1) = -1 + 3 = 2
→ (-1,2)

B(4,5):
x' = 4/2 = 2
y' = -1 + (1/2)(6) = -1 + 3 = 2
→ (2,2)

C(4,1):
x' = 4/2 = 2
y' = -1 + (1/2)(1+1) = -1 + 1 = 0
→ (2,0)

So after enlargement: (-1,2), (2,2), (2,0)

Now, reflect in the line y = -x.

Reflection over y = -x: the rule is (x,y) → (-y, -x)

Because swapping and negating.

Standard: reflection over y=-x maps (x,y) to (-y, -x)

So:

(-1,2) → (-2, 1) [ -y = -2, -x = -(-1)=1 ]
(2,2) → (-2, -2)
(2,0) → (0, -2)

So Triangle J has vertices: (-2,1), (-2,-2), (0,-2)

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Problem 6: Shape K translated by vector (0,-4), then rotated 90° counterclockwise about (-1,0) to give Shape L

First, vertices of Shape K.

From grid: it’s a kite or diamond shape.

Vertices:
- (-3,3) — top?
- (-1,5) — top right?
- (1,3) — right?
- (-1,1) — bottom?

Looking: typically, for such shape, points are:

A(-3,3), B(-1,5), C(1,3), D(-1,1)

Yes, symmetric.

So vertices: (-3,3), (-1,5), (1,3), (-1,1)

First, translate by vector (0,-4): add 0 to x, -4 to y.

So:

(-3,3) → (-3,-1)
(-1,5) → (-1,1)
(1,3) → (1,-1)
(-1,1) → (-1,-3)

Now, rotate 90° counterclockwise about point (-1,0).

Rule for 90° CCW about (a,b):

First, translate so center is origin: subtract (a,b)
Then rotate 90° CCW: (x,y) → (-y, x)
Then translate back: add (a,b)

So overall:
x' = a - (y - b)
y' = b + (x - a)

Derivation:

Let (x,y) be point.

Relative to center: (x - a, y - b)

Rotate 90° CCW: (-(y - b), x - a)

Then absolute: (a - (y - b), b + (x - a))

Simplify:
x' = a - y + b
y' = b + x - a

With a = -1, b = 0

So:
x' = -1 - y + 0 = -1 - y
y' = 0 + x - (-1) = x + 1

So formula: (x,y) → (-1 - y, x + 1)

Apply to each translated point:

First point: (-3,-1)
x' = -1 - (-1) = -1 +1 = 0
y' = -3 + 1 = -2
→ (0,-2)

Second: (-1,1)
x' = -1 - 1 = -2
y' = -1 + 1 = 0
→ (-2,0)

Third: (1,-1)
x' = -1 - (-1) = 0
y' = 1 + 1 = 2
→ (0,2)

Fourth: (-1,-3)
x' = -1 - (-3) = -1 +3 = 2
y' = -1 + 1 = 0
→ (2,0)

So Shape L has vertices: (0,-2), (-2,0), (0,2), (2,0)

Which is a diamond centered at origin, but shifted? Points at (0,-2), (-2,0), (0,2), (2,0) — yes, a square rotated 45°, but actually a diamond with those vertices.

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Now, for the Final Answer, since the task is to complete the transformations by drawing, but we can't draw here, perhaps we list the final coordinates for each shape as the answer.

Since the user might expect the description, but in the context, probably the final positions are what matter.

But to follow the format, we'll box the key results.

However, the instruction is to provide final answer after reasoning, and since there are six parts, perhaps we summarize.

But let's see the response format: it says "Final Answer: [Provide only the final correct answer]"

Probably for each problem, the final shape's position.

Since it's a worksheet, and we solved it, we can state:

For Problem 1: Shape B is located with vertices at (3,-2), (1,-2), (1,-3), (1,-5), (2,-5), (2,-3), (3,-3)

But that's long.

Perhaps in the context, the answer is the drawing, but since we can't, maybe we skip boxing, but the instruction requires it.

Another idea: perhaps the "final answer" is not applicable since it's a drawing task, but the user said "solve the problem accurately", so we did the math.

To comply, I'll box the final coordinates for each shape as a summary.

But let's make it concise.

Since the assistant is to help student, and the student needs to draw, perhaps the answer is the steps, but the format requires a final answer box.

I think for such tasks, the final answer is the description of the final shape's location.

So for each problem, we can say:

1) Shape B: vertices at (1,-2), (1,-3), (1,-5), (2,-5), (2,-3), (3,-3), (3,-2) — but ordered properly.

From earlier calculation for Problem 1, we have the points.

To save space, perhaps list the min and max or something, but better to list key points.

Since the problem is to draw, and we've calculated, for the Final Answer section, we can write:

After performing all transformations, the final shapes are positioned as follows:

- Shape B: around (1,-2) to (3,-2) etc.
But let's compile.

Perhaps the expected final answer is not numerical, but since the instruction is to box, and for math problems, we box the answer, here it's geometric.

I recall that in some systems, for such worksheets, the answer is the drawing, but since we're text-based, we'll provide the coordinates.

So for Final Answer, I'll list the vertices for each final shape.

But to keep it neat:

Final Answer for each problem:

1) Shape B vertices: (1,-2), (1,-3), (1,-5), (2,-5), (2,-3), (3,-3), (3,-2)

2) Triangle D vertices: (1,-2), (4,-2), (4,-6)

3) Shape F vertices: (0,4), (-4,4), (-4,2), (-2,2), (-2,0), (0,0), (0,2)

4) Shape H vertices: (-3,4), (-3,2), (-1,1), (-1,4)

5) Triangle J vertices: (-2,1), (-2,-2), (0,-2)

6) Shape L vertices: (0,-2), (-2,0), (0,2), (2,0)

Now, for the box, perhaps we can write it as a list.

Since the format might expect a single box, but I think it's fine.

Or perhaps the final answer is the completion of the drawings, but we can't do that.

Another thought: in the context of the user, they might want the answers to check their work, so providing the final coordinates is helpful.

So I'll proceed.

But let's double-check one for accuracy.

For example, Problem 2: Triangle C vertices (0,3), (0,0), (4,0)

Rotate 180° about (-1,2):

(0,3) -> (2*(-1)-0, 2*2-3) = (-2,1)

(0,0) -> (-2,4)

(4,0) -> (2*(-1)-4, 4-0) = (-6,4)

Reflect over y=x: swap x and y: (-2,1)->(1,-2), (-2,4)->(4,-2), (-6,4)->(4,-6) — yes.

Similarly, others seem correct.

So for Final Answer, we'll box the list.

But to fit the format, perhaps write:

The final positions of the transformed shapes are:

1) Shape B: vertices at (1,-2), (1,-3), (1,-5), (2,-5), (2,-3), (3,-3), (3,-2)

2) Triangle D: vertices at (1,-2), (4,-2), (4,-6)

3) Shape F: vertices at (0,4), (-4,4), (-4,2), (-2,2), (-2,0), (0,0), (0,2)

4) Shape H: vertices at (-3,4), (-3,2), (-1,1), (-1,4)

5) Triangle J: vertices at (-2,1), (-2,-2), (0,-2)

6) Shape L: vertices at (0,-2), (-2,0), (0,2), (2,0)

Now, for the box, since it's multiple, perhaps box the entire thing, but usually box is for single answer.

Given that, and since the problem is to draw, perhaps the final answer is not boxed, but the instruction says "provide only the final correct answer" after reasoning.

I think for this case, we can consider the final answer as the set of descriptions.

To comply, I'll box the summary.

Perhaps the user expects the answer for each, but in the response, we can state.

Another idea: in some platforms, for such tasks, the answer is the drawing, but since we're AI, we provide the coordinates.

I think it's acceptable.

So for the Final Answer section, I'll write the list.

But to make it clean, let's write:

After completing all transformations, the final shapes have the following vertex coordinates:

- Problem 1 (Shape B): (1,-2), (1,-3), (1,-5), (2,-5), (2,-3), (3,-3), (3,-2)
- Problem 2 (Triangle D): (1,-2), (4,-2), (4,-6)
- Problem 3 (Shape F): (0,4), (-4,4), (-4,2), (-2,2), (-2,0), (0,0), (0,2)
- Problem 4 (Shape H): (-3,4), (-3,2), (-1,1), (-1,4)
- Problem 5 (Triangle J): (-2,1), (-2,-2), (0,-2)
- Problem 6 (Shape L): (0,-2), (-2,0), (0,2), (2,0)

And box this text.

But typically, boxing is for mathematical expressions.

Perhaps for each problem, the answer is implied, but I think this is fine.

Since the assistant is to help, and the student can use this to verify their drawing, it should be ok.

So I'll proceed.