Section A of the Combining Transformations worksheet featuring six problems requiring students to draw shapes after multiple geometric changes.

Let’s solve each problem step by step. We’ll go one at a time, carefully tracking the transformations.

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

Shape A is an L-shape with corners at:
- Top-left: (-5, 5)
- Top-right: (-3, 5)
- Bottom-right: (-3, 1)
- Bottom-left (inner corner): (-4, 1)
- Then down to (-4, 3)? Wait — let’s list all vertices clearly.

Actually, looking at the grid:

Shape A has these key points (going clockwise from top-left):
- (-5, 5)
- (-3, 5)
- (-3, 1)
- (-4, 1)
- (-4, 3)
- (-5, 3)
→ back to (-5, 5)

But actually, it's simpler: it’s a rectangle from x=-5 to x=-3 and y=3 to y=5, plus a rectangle below from x=-4 to x=-3 and y=1 to y=3? No — better to plot the outline.

Actually, from the image description, Shape A is made of two rectangles:
- Left part: from x=-5 to x=-4, y=3 to y=5 → width 1, height 2
- Right part: from x=-4 to x=-3, y=1 to y=5 → width 1, height 4

So vertices are:
(-5,5), (-3,5), (-3,1), (-4,1), (-4,3), (-5,3)

Now, reflect over y-axis: change sign of x-coordinate.

Reflected points:
(5,5), (3,5), (3,1), (4,1), (4,3), (5,3)

Now translate by vector (-2, -6): subtract 2 from x, subtract 6 from y.

New points:
(5-2, 5-6) = (3, -1)
(3-2, 5-6) = (1, -1)
(3-2, 1-6) = (1, -5)
(4-2, 1-6) = (2, -5)
(4-2, 3-6) = (2, -3)
(5-2, 3-6) = (3, -3)

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

We can draw this on the grid — it’s an L-shape pointing down-left, starting at (1,-1) going right to (3,-1), down to (3,-3), left to (2,-3), down to (2,-5), left to (1,-5), up to (1,-1).

✔ Done for Problem 1.

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

Triangle C has vertices:
From grid: (0,3), (0,0), (4,0)

Step 1: Rotate 180° about point (-1,2)

Rule for 180° rotation about (a,b):
(x,y) → (2a - x, 2b - y)

So for each point:

Point (0,3):
x’ = 2*(-1) - 0 = -2
y’ = 2*2 - 3 = 4 - 3 = 1
→ (-2, 1)

Point (0,0):
x’ = -2 - 0 = -2
y’ = 4 - 0 = 4
→ (-2, 4)

Point (4,0):
x’ = -2 - 4 = -6
y’ = 4 - 0 = 4
→ (-6, 4)

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

Step 2: Reflect in line y = x → swap x and y coordinates.

So:

(-2,1) → (1, -2)
(-2,4) → (4, -2)
(-6,4) → (4, -6)

Wait — that doesn’t look right. Let me double-check.

Reflection over y=x: (x,y) → (y,x)

So:

(-2,1) → (1, -2) ✔
(-2,4) → (4, -2) ✔
(-6,4) → (4, -6) ✔

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

That’s a right triangle with right angle at (4,-2), going left to (1,-2) and down to (4,-6).

✔ Done for Problem 2.

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

First, identify Shape E.

From grid: It’s a T-shape or cross? Actually, looks like:

Bottom row: from x=1 to x=3, y=-6 → so points (1,-6), (2,-6), (3,-6)
Then above: (2,-5), (2,-4), (2,-3) — wait no.

Looking again: Shape E is drawn as:

- Horizontal bar: from x=1 to x=3 at y=-6 → three squares wide
- Vertical stem: from y=-5 to y=-3 at x=2 → three squares tall

So vertices (outline):

Start at (1,-6), go to (3,-6), up to (3,-5), left to (2,-5), up to (2,-3), left to (1,-3), down to (1,-5), right to (1,-6)? That’s messy.

Better to take key corners:

Top of vertical: (2,-3)
Bottom of vertical: (2,-6) — but connected to horizontal.

Actually, the shape occupies:

Cells:
(1,-6), (2,-6), (3,-6) — bottom row
(2,-5), (2,-4), (2,-3) — middle column

So outer boundary vertices (clockwise from top):

(2,-3) → (3,-3)? No, only (2,-3) is top.

Actually, since it’s blocky, we can use the four corners of the bounding box plus indentations, but for enlargement, we need actual vertex points.

Let’s define the polygon vertices in order:

Start at (1,-6) → (3,-6) → (3,-5) → (2,-5) → (2,-3) → (1,-3) → (1,-5) → back to (1,-6)

Yes, that traces the outline.

So vertices:
A: (1,-6)
B: (3,-6)
C: (3,-5)
D: (2,-5)
E: (2,-3)
F: (1,-3)
G: (1,-5)
Back to A: (1,-6)

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

Rule for enlargement with negative scale factor:
For any point P, vector from center O to P is multiplied by k, then added to O.

Formula:
P' = O + k*(P - O)

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

So for each point:

Take point A: (1,-6)

Vector from O to A: (1-2, -6-(-3)) = (-1, -3)

Multiply by -2: (2, 6)

Add to O: (2+2, -3+6) = (4, 3)

Point B: (3,-6)

Vector: (3-2, -6+3) = (1, -3)

× -2: (-2, 6)

Add to O: (2-2, -3+6) = (0, 3)

Point C: (3,-5)

Vector: (1, -2)

× -2: (-2, 4)

Add to O: (0, 1)

Point D: (2,-5)

Vector: (0, -2)

× -2: (0, 4)

Add to O: (2, 1)

Point E: (2,-3) → same as center → stays (2,-3)? Wait:

Vector: (0,0) × -2 = (0,0) → add to O → (2,-3)

But since it’s the center, it should stay fixed? Yes.

Point F: (1,-3)

Vector: (-1, 0)

× -2: (2, 0)

Add to O: (4, -3)

Point G: (1,-5)

Vector: (-1, -2)

× -2: (2, 4)

Add to O: (4, 1)

So after enlargement, new points:

A': (4,3)
B': (0,3)
C': (0,1)
D': (2,1)
E': (2,-3)
F': (4,-3)
G': (4,1)

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

Final points for Shape F:

A'': (4-4, 3+3) = (0,6)
B'': (0-4, 3+3) = (-4,6)
C'': (0-4, 1+3) = (-4,4)
D'': (2-4, 1+3) = (-2,4)
E'': (2-4, -3+3) = (-2,0)
F'': (4-4, -3+3) = (0,0)
G'': (4-4, 1+3) = (0,4)

So Shape F has vertices:
(0,6), (-4,6), (-4,4), (-2,4), (-2,0), (0,0), (0,4)

Plotting this: it’s a larger version of the original T-shape, flipped and moved.

✔ Done for Problem 3.

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Problem 4: Shape G → rotated 90° clockwise about origin → then reflected in line x = -3 → Shape H

Shape G: from grid, it’s a trapezoid or irregular quadrilateral.

Vertices (from image):

Assume:
Left-bottom: (-5,-5)
Right-bottom: (-2,-5)
Right-top: (-3,-3)
Left-top: (-5,-3)

Wait — let’s check:

It looks like a parallelogram or trapezoid.

Points:
(-5,-5), (-2,-5), (-3,-3), (-5,-3)

Yes — that makes sense.

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

Step 1: Rotate 90° clockwise about origin (0,0)

Rule: (x,y) → (y, -x)

So:

A: (-5,-5) → (-5, 5)
B: (-2,-5) → (-5, 2)
C: (-3,-3) → (-3, 3)
D: (-5,-3) → (-3, 5)

After rotation:
A': (-5,5)
B': (-5,2)
C': (-3,3)
D': (-3,5)

Step 2: Reflect in line x = -3

Rule for reflection over vertical line x = a:
x' = 2a - x
y' = y

Here a = -3

So for each point:

A': (-5,5) → x' = 2*(-3) - (-5) = -6 + 5 = -1 → (-1,5)
B': (-5,2) → x' = -6 +5 = -1 → (-1,2)
C': (-3,3) → x' = -6 - (-3) = -3 → (-3,3)
D': (-3,5) → x' = -6 - (-3) = -3 → (-3,5)

So Shape H has vertices:
(-1,5), (-1,2), (-3,3), (-3,5)

This is a quadrilateral — we can connect them in order.

✔ Done for Problem 4.

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

Triangle I vertices: from grid

Looks like:
(-2,5), (4,5), (4,1)

Confirm: yes, right triangle with right angle at (4,5)? No — from (-2,5) to (4,5) to (4,1) — so right angle at (4,5)? Wait, no:

Points:
A: (-2,5)
B: (4,5)
C: (4,1)

So AB horizontal, BC vertical — right angle at B (4,5)

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

Formula: P' = O + k*(P - O)

O = (0,-1), k = 0.5

Point A: (-2,5)

Vector: (-2-0, 5-(-1)) = (-2,6)

× 0.5: (-1,3)

Add to O: (0-1, -1+3) = (-1,2)

Point B: (4,5)

Vector: (4,6)

× 0.5: (2,3)

Add to O: (2,2)

Point C: (4,1)

Vector: (4,2)

× 0.5: (2,1)

Add to O: (2,0)

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

Step 2: Reflect in line y = -x

Rule: (x,y) → (-y, -x)

So:

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

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

✔ Done for Problem 5.

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Problem 6: Shape K → translated by vector (0,-4) → then rotated 90° anti-clockwise about (-1,0) → Shape L

Shape K: diamond/rhombus

Vertices: from grid

Top: (-2,6)
Right: (-1,3)
Bottom: (-2,0)
Left: (-3,3)

Check: yes, symmetric.

So vertices:
A: (-2,6)
B: (-1,3)
C: (-2,0)
D: (-3,3)

Step 1: Translate by vector (0,-4) → add 0 to x, subtract 4 from y

A': (-2,2)
B': (-1,-1)
C': (-2,-4)
D': (-3,-1)

Step 2: Rotate 90° anti-clockwise about point (-1,0)

Rule for 90° CCW about (a,b):
(x,y) → (a - (y - b), b + (x - a))

Or: relative to center: (dx, dy) = (x-a, y-b)
Rotate 90° CCW: (-dy, dx)
Then add back: (a - dy, b + dx)

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

With a = -1, b = 0

Apply to each point:

Point A': (-2,2)

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

Point B': (-1,-1)

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

Point C': (-2,-4)

x' = -1 - (-4 - 0) = -1 - (-4) = 3
y' = 0 + (-2 - (-1)) = 0 + (-1) = -1
→ (3, -1)

Point D': (-3,-1)

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

So Shape L has vertices:
(-3,-1), (0,0), (3,-1), (0,-2)

Plotting: it’s a kite shape centered around x=0, y=-1.

✔ Done for Problem 6.

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Now, since the task was to “complete the transformations by drawing”, but we’re text-based, we’ve computed the final coordinates for each transformed shape.

The student would draw these on their grids using the coordinates we found.

Since the user didn’t ask for explanation or teaching, and GeneratePractice is false, we just provide the final answers as the set of vertices for each resulting shape.

Final Answer:

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