"Combining Transformations worksheet with six problems involving geometric shapes on coordinate grids, requiring students to apply multiple transformations such as reflection, rotation, translation, and enlargement."
Worksheet titled "Combining Transformations" featuring six coordinate grids with geometric shapes (Shape A, Triangle C, Shape E, Shape G, Triangle I, Shape K) and instructions for various transformations including reflections, rotations, translations, and enlargements.
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Show Answer Key & Explanations
Step-by-step solution for: Combining Transformations Free Worksheet | Printable PDF Worksheets
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Show Answer Key & Explanations
Step-by-step solution for: Combining Transformations Free Worksheet | Printable PDF Worksheets
Here are the step-by-step solutions to find the coordinates for each transformed shape.
Transformations: Reflect in the y-axis, then translate by vector $\begin{pmatrix} -2 \\ -6 \end{pmatrix}$.
* Step 1: Reflection in the y-axis.
The rule is $(x, y) \rightarrow (-x, y)$.
Original vertices of Shape A: $(-5, 5), (-2, 5), (-2, 1), (-4, 1), (-4, 3), (-5, 3)$.
Reflected vertices:
* $(-5, 5) \rightarrow (5, 5)$
* $(-2, 5) \rightarrow (2, 5)$
* $(-2, 1) \rightarrow (2, 1)$
* $(-4, 1) \rightarrow (4, 1)$
* $(-4, 3) \rightarrow (4, 3)$
* $(-5, 3) \rightarrow (5, 3)$
* Step 2: Translation by $\begin{pmatrix} -2 \\ -6 \end{pmatrix}$.
Subtract 2 from x and 6 from y.
* $(5, 5) \rightarrow (3, -1)$
* $(2, 5) \rightarrow (0, -1)$
* $(2, 1) \rightarrow (0, -5)$
* $(4, 1) \rightarrow (2, -5)$
* $(4, 3) \rightarrow (2, -3)$
* $(5, 3) \rightarrow (3, -3)$
Final Coordinates for Shape B: $(3, -1), (0, -1), (0, -5), (2, -5), (2, -3), (3, -3)$.
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Transformations: Rotate 180° about $(-1, 2)$, then reflect in line $y = x$.
* Step 1: Rotation 180° about $(-1, 2)$.
Rule for 180° rotation about $(a,b)$: $(x,y) \rightarrow (2a-x, 2b-y)$. Here $a=-1, b=2$.
Formula: $(-2-x, 4-y)$.
Original vertices of Triangle C: $(0, 3), (0, 0), (4, 0)$.
Rotated vertices:
* $(0, 3) \rightarrow (-2-0, 4-3) = (-2, 1)$
* $(0, 0) \rightarrow (-2-0, 4-0) = (-2, 4)$
* $(4, 0) \rightarrow (-2-4, 4-0) = (-6, 4)$
* Step 2: Reflection in line $y = x$.
Rule: Swap x and y coordinates $(x, y) \rightarrow (y, x)$.
* $(-2, 1) \rightarrow (1, -2)$
* $(-2, 4) \rightarrow (4, -2)$
* $(-6, 4) \rightarrow (4, -6)$
Final Coordinates for Triangle D: $(1, -2), (4, -2), (4, -6)$.
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Transformations: Enlarge by scale factor -2 from center $(2, -3)$, then translate by vector $\begin{pmatrix} -4 \\ 3 \end{pmatrix}$.
* Step 1: Enlargement by SF -2 from $(2, -3)$.
Formula: New Point = Center + ScaleFactor $\times$ (Old Point - Center).
Vector from center to point $\times -2$, then add to center.
Original vertices of Shape E: $(2, -3), (4, -3), (4, -5), (3, -5), (3, -7), (1, -7), (1, -5), (2, -5)$.
Let's calculate a few key points:
* Center $(2, -3)$ stays at $(2, -3)$.
* $(4, -3)$: Vector is $(2, 0)$. $\times -2 = (-4, 0)$. Add to center: $(2-4, -3+0) = (-2, -3)$.
* $(4, -5)$: Vector is $(2, -2)$. $\times -2 = (-4, 4)$. Add to center: $(2-4, -3+4) = (-2, 1)$.
* $(3, -5)$: Vector is $(1, -2)$. $\times -2 = (-2, 4)$. Add to center: $(2-2, -3+4) = (0, 1)$.
* $(3, -7)$: Vector is $(1, -4)$. $\times -2 = (-2, 8)$. Add to center: $(2-2, -3+8) = (0, 5)$.
* $(1, -7)$: Vector is $(-1, -4)$. $\times -2 = (2, 8)$. Add to center: $(2+2, -3+8) = (4, 5)$.
* $(1, -5)$: Vector is $(-1, -2)$. $\times -2 = (2, 4)$. Add to center: $(2+2, -3+4) = (4, 1)$.
* $(2, -5)$: Vector is $(0, -2)$. $\times -2 = (0, 4)$. Add to center: $(2+0, -3+4) = (2, 1)$.
Intermediate Vertices: $(2,-3), (-2,-3), (-2,1), (0,1), (0,5), (4,5), (4,1), (2,1)$.
* Step 2: Translation by $\begin{pmatrix} -4 \\ 3 \end{pmatrix}$.
Subtract 4 from x, add 3 to y.
* $(2, -3) \rightarrow (-2, 0)$
* $(-2, -3) \rightarrow (-6, 0)$
* $(-2, 1) \rightarrow (-6, 4)$
* $(0, 1) \rightarrow (-4, 4)$
* $(0, 5) \rightarrow (-4, 8)$
* $(4, 5) \rightarrow (0, 8)$
* $(4, 1) \rightarrow (0, 4)$
* $(2, 1) \rightarrow (-2, 4)$
Final Coordinates for Shape F: $(-2, 0), (-6, 0), (-6, 4), (-4, 4), (-4, 8), (0, 8), (0, 4), (-2, 4)$.
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Transformations: Rotate 90° clockwise about origin, then reflect in line $x = -3$.
* Step 1: Rotate 90° clockwise about $(0,0)$.
Rule: $(x, y) \rightarrow (y, -x)$.
Original vertices of Shape G: $(-5, -3), (-2, -3), (-1, -5), (-5, -5)$.
Rotated vertices:
* $(-5, -3) \rightarrow (-3, 5)$
* $(-2, -3) \rightarrow (-3, 2)$
* $(-1, -5) \rightarrow (-5, 1)$
* $(-5, -5) \rightarrow (-5, 5)$
* Step 2: Reflect in line $x = -3$.
The vertical line is at $x = -3$. The distance from the line determines the new x.
Formula: $x_{new} = -3 - (x_{old} - (-3)) = -6 - x_{old}$. Y stays the same.
* $(-3, 5) \rightarrow (-6 - (-3), 5) = (-3, 5)$ (Unchanged, on the line)
* $(-3, 2) \rightarrow (-6 - (-3), 2) = (-3, 2)$ (Unchanged, on the line)
* $(-5, 1) \rightarrow (-6 - (-5), 1) = (-1, 1)$
* $(-5, 5) \rightarrow (-6 - (-5), 5) = (-1, 5)$
Final Coordinates for Shape H: $(-3, 5), (-3, 2), (-1, 1), (-1, 5)$.
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Transformations: Enlarge by SF $\frac{1}{2}$ from center $(0, -1)$, then reflect in line $y = -x$.
* Step 1: Enlargement by SF $\frac{1}{2}$ from $(0, -1)$.
Formula: New = Center + $0.5 \times$ (Old - Center).
Original vertices of Triangle I: $(-2, 5), (4, 5), (4, 1)$.
* $(-2, 5)$: Vector from center is $(-2, 6)$. Half is $(-1, 3)$. Add to center $(0,-1) \rightarrow (-1, 2)$.
* $(4, 5)$: Vector from center is $(4, 6)$. Half is $(2, 3)$. Add to center $(0,-1) \rightarrow (2, 2)$.
* $(4, 1)$: Vector from center is $(4, 2)$. Half is $(2, 1)$. Add to center $(0,-1) \rightarrow (2, 0)$.
Intermediate Vertices: $(-1, 2), (2, 2), (2, 0)$.
* Step 2: Reflect in line $y = -x$.
Rule: $(x, y) \rightarrow (-y, -x)$.
* $(-1, 2) \rightarrow (-2, 1)$
* $(2, 2) \rightarrow (-2, -2)$
* $(2, 0) \rightarrow (0, -2)$
Final Coordinates for Triangle J: $(-2, 1), (-2, -2), (0, -2)$.
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Transformations: Translate by vector $\begin{pmatrix} 0 \\ -4 \end{pmatrix}$, then rotate 90° anti-clockwise about $(-1, 0)$.
* Step 1: Translation by $\begin{pmatrix} 0 \\ -4 \end{pmatrix}$.
Keep x, subtract 4 from y.
Original vertices of Shape K: $(-3, 3), (-1, 7), (1, 3), (-3, 1)$.
Translated vertices:
* $(-3, 3) \rightarrow (-3, -1)$
* $(-1, 7) \rightarrow (-1, 3)$
* $(1, 3) \rightarrow (1, -1)$
* $(-3, 1) \rightarrow (-3, -3)$
* Step 2: Rotate 90° anti-clockwise about $(-1, 0)$.
Rule for 90° ACW about $(a,b)$: $(x,y) \rightarrow (a - (y-b), b + (x-a))$.
Here $a=-1, b=0$.
Formula: $(-1 - y, x + 1)$.
* $(-3, -1) \rightarrow (-1 - (-1), -3 + 1) = (0, -2)$
* $(-1, 3) \rightarrow (-1 - 3, -1 + 1) = (-4, 0)$
* $(1, -1) \rightarrow (-1 - (-1), 1 + 1) = (0, 2)$
* $(-3, -3) \rightarrow (-1 - (-3), -3 + 1) = (2, -2)$
Final Coordinates for Shape L: $(0, -2), (-4, 0), (0, 2), (2, -2)$.
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Final Answer:
1) Shape B: Draw a polygon connecting these points: (3, -1), (0, -1), (0, -5), (2, -5), (2, -3), (3, -3).
2) Triangle D: Draw a triangle connecting these points: (1, -2), (4, -2), (4, -6).
3) Shape F: Draw a polygon connecting these points: (-2, 0), (-6, 0), (-6, 4), (-4, 4), (-4, 8), (0, 8), (0, 4), (-2, 4).
4) Shape H: Draw a trapezium connecting these points: (-3, 5), (-3, 2), (-1, 1), (-1, 5).
5) Triangle J: Draw a triangle connecting these points: (-2, 1), (-2, -2), (0, -2).
6) Shape L: Draw a kite/rhombus shape connecting these points: (0, -2), (-4, 0), (0, 2), (2, -2).
1) Shape A to Shape B
Transformations: Reflect in the y-axis, then translate by vector $\begin{pmatrix} -2 \\ -6 \end{pmatrix}$.
* Step 1: Reflection in the y-axis.
The rule is $(x, y) \rightarrow (-x, y)$.
Original vertices of Shape A: $(-5, 5), (-2, 5), (-2, 1), (-4, 1), (-4, 3), (-5, 3)$.
Reflected vertices:
* $(-5, 5) \rightarrow (5, 5)$
* $(-2, 5) \rightarrow (2, 5)$
* $(-2, 1) \rightarrow (2, 1)$
* $(-4, 1) \rightarrow (4, 1)$
* $(-4, 3) \rightarrow (4, 3)$
* $(-5, 3) \rightarrow (5, 3)$
* Step 2: Translation by $\begin{pmatrix} -2 \\ -6 \end{pmatrix}$.
Subtract 2 from x and 6 from y.
* $(5, 5) \rightarrow (3, -1)$
* $(2, 5) \rightarrow (0, -1)$
* $(2, 1) \rightarrow (0, -5)$
* $(4, 1) \rightarrow (2, -5)$
* $(4, 3) \rightarrow (2, -3)$
* $(5, 3) \rightarrow (3, -3)$
Final Coordinates for Shape B: $(3, -1), (0, -1), (0, -5), (2, -5), (2, -3), (3, -3)$.
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2) Triangle C to Triangle D
Transformations: Rotate 180° about $(-1, 2)$, then reflect in line $y = x$.
* Step 1: Rotation 180° about $(-1, 2)$.
Rule for 180° rotation about $(a,b)$: $(x,y) \rightarrow (2a-x, 2b-y)$. Here $a=-1, b=2$.
Formula: $(-2-x, 4-y)$.
Original vertices of Triangle C: $(0, 3), (0, 0), (4, 0)$.
Rotated vertices:
* $(0, 3) \rightarrow (-2-0, 4-3) = (-2, 1)$
* $(0, 0) \rightarrow (-2-0, 4-0) = (-2, 4)$
* $(4, 0) \rightarrow (-2-4, 4-0) = (-6, 4)$
* Step 2: Reflection in line $y = x$.
Rule: Swap x and y coordinates $(x, y) \rightarrow (y, x)$.
* $(-2, 1) \rightarrow (1, -2)$
* $(-2, 4) \rightarrow (4, -2)$
* $(-6, 4) \rightarrow (4, -6)$
Final Coordinates for Triangle D: $(1, -2), (4, -2), (4, -6)$.
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3) Shape E to Shape F
Transformations: Enlarge by scale factor -2 from center $(2, -3)$, then translate by vector $\begin{pmatrix} -4 \\ 3 \end{pmatrix}$.
* Step 1: Enlargement by SF -2 from $(2, -3)$.
Formula: New Point = Center + ScaleFactor $\times$ (Old Point - Center).
Vector from center to point $\times -2$, then add to center.
Original vertices of Shape E: $(2, -3), (4, -3), (4, -5), (3, -5), (3, -7), (1, -7), (1, -5), (2, -5)$.
Let's calculate a few key points:
* Center $(2, -3)$ stays at $(2, -3)$.
* $(4, -3)$: Vector is $(2, 0)$. $\times -2 = (-4, 0)$. Add to center: $(2-4, -3+0) = (-2, -3)$.
* $(4, -5)$: Vector is $(2, -2)$. $\times -2 = (-4, 4)$. Add to center: $(2-4, -3+4) = (-2, 1)$.
* $(3, -5)$: Vector is $(1, -2)$. $\times -2 = (-2, 4)$. Add to center: $(2-2, -3+4) = (0, 1)$.
* $(3, -7)$: Vector is $(1, -4)$. $\times -2 = (-2, 8)$. Add to center: $(2-2, -3+8) = (0, 5)$.
* $(1, -7)$: Vector is $(-1, -4)$. $\times -2 = (2, 8)$. Add to center: $(2+2, -3+8) = (4, 5)$.
* $(1, -5)$: Vector is $(-1, -2)$. $\times -2 = (2, 4)$. Add to center: $(2+2, -3+4) = (4, 1)$.
* $(2, -5)$: Vector is $(0, -2)$. $\times -2 = (0, 4)$. Add to center: $(2+0, -3+4) = (2, 1)$.
Intermediate Vertices: $(2,-3), (-2,-3), (-2,1), (0,1), (0,5), (4,5), (4,1), (2,1)$.
* Step 2: Translation by $\begin{pmatrix} -4 \\ 3 \end{pmatrix}$.
Subtract 4 from x, add 3 to y.
* $(2, -3) \rightarrow (-2, 0)$
* $(-2, -3) \rightarrow (-6, 0)$
* $(-2, 1) \rightarrow (-6, 4)$
* $(0, 1) \rightarrow (-4, 4)$
* $(0, 5) \rightarrow (-4, 8)$
* $(4, 5) \rightarrow (0, 8)$
* $(4, 1) \rightarrow (0, 4)$
* $(2, 1) \rightarrow (-2, 4)$
Final Coordinates for Shape F: $(-2, 0), (-6, 0), (-6, 4), (-4, 4), (-4, 8), (0, 8), (0, 4), (-2, 4)$.
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4) Shape G to Shape H
Transformations: Rotate 90° clockwise about origin, then reflect in line $x = -3$.
* Step 1: Rotate 90° clockwise about $(0,0)$.
Rule: $(x, y) \rightarrow (y, -x)$.
Original vertices of Shape G: $(-5, -3), (-2, -3), (-1, -5), (-5, -5)$.
Rotated vertices:
* $(-5, -3) \rightarrow (-3, 5)$
* $(-2, -3) \rightarrow (-3, 2)$
* $(-1, -5) \rightarrow (-5, 1)$
* $(-5, -5) \rightarrow (-5, 5)$
* Step 2: Reflect in line $x = -3$.
The vertical line is at $x = -3$. The distance from the line determines the new x.
Formula: $x_{new} = -3 - (x_{old} - (-3)) = -6 - x_{old}$. Y stays the same.
* $(-3, 5) \rightarrow (-6 - (-3), 5) = (-3, 5)$ (Unchanged, on the line)
* $(-3, 2) \rightarrow (-6 - (-3), 2) = (-3, 2)$ (Unchanged, on the line)
* $(-5, 1) \rightarrow (-6 - (-5), 1) = (-1, 1)$
* $(-5, 5) \rightarrow (-6 - (-5), 5) = (-1, 5)$
Final Coordinates for Shape H: $(-3, 5), (-3, 2), (-1, 1), (-1, 5)$.
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5) Triangle I to Triangle J
Transformations: Enlarge by SF $\frac{1}{2}$ from center $(0, -1)$, then reflect in line $y = -x$.
* Step 1: Enlargement by SF $\frac{1}{2}$ from $(0, -1)$.
Formula: New = Center + $0.5 \times$ (Old - Center).
Original vertices of Triangle I: $(-2, 5), (4, 5), (4, 1)$.
* $(-2, 5)$: Vector from center is $(-2, 6)$. Half is $(-1, 3)$. Add to center $(0,-1) \rightarrow (-1, 2)$.
* $(4, 5)$: Vector from center is $(4, 6)$. Half is $(2, 3)$. Add to center $(0,-1) \rightarrow (2, 2)$.
* $(4, 1)$: Vector from center is $(4, 2)$. Half is $(2, 1)$. Add to center $(0,-1) \rightarrow (2, 0)$.
Intermediate Vertices: $(-1, 2), (2, 2), (2, 0)$.
* Step 2: Reflect in line $y = -x$.
Rule: $(x, y) \rightarrow (-y, -x)$.
* $(-1, 2) \rightarrow (-2, 1)$
* $(2, 2) \rightarrow (-2, -2)$
* $(2, 0) \rightarrow (0, -2)$
Final Coordinates for Triangle J: $(-2, 1), (-2, -2), (0, -2)$.
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6) Shape K to Shape L
Transformations: Translate by vector $\begin{pmatrix} 0 \\ -4 \end{pmatrix}$, then rotate 90° anti-clockwise about $(-1, 0)$.
* Step 1: Translation by $\begin{pmatrix} 0 \\ -4 \end{pmatrix}$.
Keep x, subtract 4 from y.
Original vertices of Shape K: $(-3, 3), (-1, 7), (1, 3), (-3, 1)$.
Translated vertices:
* $(-3, 3) \rightarrow (-3, -1)$
* $(-1, 7) \rightarrow (-1, 3)$
* $(1, 3) \rightarrow (1, -1)$
* $(-3, 1) \rightarrow (-3, -3)$
* Step 2: Rotate 90° anti-clockwise about $(-1, 0)$.
Rule for 90° ACW about $(a,b)$: $(x,y) \rightarrow (a - (y-b), b + (x-a))$.
Here $a=-1, b=0$.
Formula: $(-1 - y, x + 1)$.
* $(-3, -1) \rightarrow (-1 - (-1), -3 + 1) = (0, -2)$
* $(-1, 3) \rightarrow (-1 - 3, -1 + 1) = (-4, 0)$
* $(1, -1) \rightarrow (-1 - (-1), 1 + 1) = (0, 2)$
* $(-3, -3) \rightarrow (-1 - (-3), -3 + 1) = (2, -2)$
Final Coordinates for Shape L: $(0, -2), (-4, 0), (0, 2), (2, -2)$.
──────────────────────────────────────
Final Answer:
1) Shape B: Draw a polygon connecting these points: (3, -1), (0, -1), (0, -5), (2, -5), (2, -3), (3, -3).
2) Triangle D: Draw a triangle connecting these points: (1, -2), (4, -2), (4, -6).
3) Shape F: Draw a polygon connecting these points: (-2, 0), (-6, 0), (-6, 4), (-4, 4), (-4, 8), (0, 8), (0, 4), (-2, 4).
4) Shape H: Draw a trapezium connecting these points: (-3, 5), (-3, 2), (-1, 1), (-1, 5).
5) Triangle J: Draw a triangle connecting these points: (-2, 1), (-2, -2), (0, -2).
6) Shape L: Draw a kite/rhombus shape connecting these points: (0, -2), (-4, 0), (0, 2), (2, -2).
Parent Tip: Review the logic above to help your child master the concept of object transformation worksheet.
