Combining Transformations Free Worksheet | Printable PDF Worksheets - Free Printable

Let's solve each of the six transformation problems step by step, explaining the process and how to draw the resulting shapes on the coordinate grids.

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1) Shape A is first reflected in the y-axis and then translated by the vector $\begin{pmatrix} -2 \\ -6 \end{pmatrix}$ to give Shape B.

Step 1: Reflect Shape A in the y-axis
- Reflection in the y-axis changes the sign of the x-coordinate: $(x, y) \rightarrow (-x, y)$
- Shape A has vertices:
- $(-5, 5), (-3, 5), (-3, 4), (-5, 4), (-5, 3), (-3, 3), (-3, 2), (-5, 2)$
- After reflection:
- $(5, 5), (3, 5), (3, 4), (5, 4), (5, 3), (3, 3), (3, 2), (5, 2)$

Step 2: Translate by vector $\begin{pmatrix} -2 \\ -6 \end{pmatrix}$
- Add $-2$ to x, $-6$ to y:
- $(5-2, 5-6) = (3, -1)$
- $(3-2, 5-6) = (1, -1)$
- $(3-2, 4-6) = (1, -2)$
- $(5-2, 4-6) = (3, -2)$
- $(5-2, 3-6) = (3, -3)$
- $(3-2, 3-6) = (1, -3)$
- $(3-2, 2-6) = (1, -4)$
- $(5-2, 2-6) = (3, -4)$

Final Shape B: Plot these points and connect them to form a mirrored and shifted version of Shape A.

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2) Triangle C is rotated 180° from the point (-1, 2) and is then reflected in the line $y = x$ to give Triangle D.

Step 1: Rotate 180° about (-1, 2)
- Rotation 180° about a point $(a,b)$: $(x,y) \rightarrow (2a - x, 2b - y)$
- Centre: $(-1, 2)$ → formula: $x' = 2(-1) - x = -2 - x$, $y' = 2(2) - y = 4 - y$

Triangle C vertices:
- $(0,3), (4,3), (0,0)$

Apply rotation:
- $(0,3)$ → $(-2 - 0, 4 - 3) = (-2, 1)$
- $(4,3)$ → $(-2 - 4, 4 - 3) = (-6, 1)$
- $(0,0)$ → $(-2 - 0, 4 - 0) = (-2, 4)$

Intermediate shape after rotation: $(-2,1), (-6,1), (-2,4)$

Step 2: Reflect in line $y = x$
- Swap x and y coordinates: $(x,y) \rightarrow (y,x)$
- $(-2,1)$ → $(1,-2)$
- $(-6,1)$ → $(1,-6)$
- $(-2,4)$ → $(4,-2)$

Final Triangle D: Plot $(1,-2), (1,-6), (4,-2)$ and connect.

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3) Shape E is first enlarged by a scale factor of -2 from the centre of enlargement (2, -3), and secondly translated by the vector $\begin{pmatrix} -4 \\ 3 \end{pmatrix}$ to give Shape F.

Step 1: Enlarge by scale factor -2 from (2, -3)

Scale factor $-2$: negative means it’s inverted and twice as big.

Formula for enlargement:
$$
(x', y') = (2 + (-2)(x - 2), -3 + (-2)(y + 3)) = (2 - 2(x - 2), -3 - 2(y + 3))
$$
Simplify:
- $x' = 2 - 2x + 4 = 6 - 2x$
- $y' = -3 - 2y - 6 = -9 - 2y$

Shape E vertices:
- $(2,-5), (3,-5), (3,-3), (2,-3), (2,-1), (1,-1), (1,-3)$

Now apply:

- $(2,-5)$ → $x' = 6 - 4 = 2$, $y' = -9 - (-10) = 1$ → $(2,1)$
- $(3,-5)$ → $6 - 6 = 0$, $-9 + 10 = 1$ → $(0,1)$
- $(3,-3)$ → $6 - 6 = 0$, $-9 + 6 = -3$ → $(0,-3)$
- $(2,-3)$ → $6 - 4 = 2$, $-9 + 6 = -3$ → $(2,-3)$
- $(2,-1)$ → $6 - 4 = 2$, $-9 + 2 = -7$ → $(2,-7)$
- $(1,-1)$ → $6 - 2 = 4$, $-9 + 2 = -7$ → $(4,-7)$
- $(1,-3)$ → $6 - 2 = 4$, $-9 + 6 = -3$ → $(4,-3)$

So enlarged shape has points:
- $(2,1), (0,1), (0,-3), (2,-3), (2,-7), (4,-7), (4,-3)$

Step 2: Translate by $\begin{pmatrix} -4 \\ 3 \end{pmatrix}$
- Subtract 4 from x, add 3 to y

New points:
- $(2-4,1+3)=(-2,4)$
- $(0-4,1+3)=(-4,4)$
- $(0-4,-3+3)=(-4,0)$
- $(2-4,-3+3)=(-2,0)$
- $(2-4,-7+3)=(-2,-4)$
- $(4-4,-7+3)=(0,-4)$
- $(4-4,-3+3)=(0,0)$

Final Shape F: Plot these points and connect appropriately.

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4) Shape G is rotated 90° clockwise about the origin and is then reflected in the line $x = -3$ to give Shape H.

Step 1: Rotate 90° clockwise about origin
- Rule: $(x,y) \rightarrow (y, -x)$

Shape G vertices:
- $(-4,-3), (-2,-3), (-2,-5), (-4,-5)$ — this is a trapezoid

Apply rotation:
- $(-4,-3)$ → $(-3, 4)$
- $(-2,-3)$ → $(-3, 2)$
- $(-2,-5)$ → $(-5, 2)$
- $(-4,-5)$ → $(-5, 4)$

Rotated shape: $(-3,4), (-3,2), (-5,2), (-5,4)$

Step 2: Reflect in $x = -3$
- Distance from point to line $x = -3$: $|x + 3|$
- Reflected x: $x' = -3 - (x + 3) = -6 - x$

So $x' = -6 - x$, y unchanged

Apply:
- $(-3,4)$ → $x' = -6 - (-3) = -3$ → $(-3,4)$ (on line, stays)
- $(-3,2)$ → same → $(-3,2)$
- $(-5,2)$ → $x' = -6 - (-5) = -1$ → $(-1,2)$
- $(-5,4)$ → $x' = -6 - (-5) = -1$ → $(-1,4)$

Final Shape H: Points: $(-3,4), (-3,2), (-1,2), (-1,4)$ — a rectangle.

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5) Triangle I is enlarged by a scale factor of $\frac{1}{2}$, centre of enlargement (0, -1), and is then reflected in the line $y = -x$ to give Triangle J.

Step 1: Enlarge by scale factor $\frac{1}{2}$ from (0, -1)

Formula:
$$
(x', y') = (0 + \frac{1}{2}(x - 0), -1 + \frac{1}{2}(y + 1)) = \left(\frac{x}{2}, -1 + \frac{y+1}{2}\right)
$$

Triangle I vertices:
- $(-2,5), (4,5), (4,1)$

Apply:
- $(-2,5)$ → $(-1, -1 + (6)/2) = (-1, -1 + 3) = (-1, 2)$
- $(4,5)$ → $(2, -1 + (6)/2) = (2, -1 + 3) = (2, 2)$
- $(4,1)$ → $(2, -1 + (2)/2) = (2, -1 + 1) = (2, 0)$

Enlarged triangle: $(-1,2), (2,2), (2,0)$

Step 2: Reflect in line $y = -x$
- Rule: $(x,y) \rightarrow (-y, -x)$

Apply:
- $(-1,2)$ → $(-2, 1)$
- $(2,2)$ → $(-2, -2)$
- $(2,0)$ → $(0, -2)$

Final Triangle J: $(-2,1), (-2,-2), (0,-2)$

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6) Shape K is first translated by the vector $\begin{pmatrix} 0 \\ 4 \end{pmatrix}$ and is secondly rotated by 90° anti-clockwise about the point (-1, 0) to give Shape L.

Step 1: Translate by $\begin{pmatrix} 0 \\ 4 \end{pmatrix}$
- Add 4 to y-coordinate

Shape K vertices:
- $(-3,3), (-1,5), (-1,1), (-3,-1)$ — diamond shape

After translation:
- $(-3,7), (-1,9), (-1,5), (-3,3)$

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

Rule: For 90° anti-clockwise about point $(a,b)$:
$$
(x,y) \rightarrow (a - (y - b), b + (x - a))
$$
Or: $(x,y)$ → $(- (y - b) + a, (x - a) + b)$

Here $a = -1, b = 0$

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

Apply to each point:

1. $(-3,7)$ → $x' = -1 - 7 = -8$, $y' = -3 + 1 = -2$ → $(-8,-2)$
2. $(-1,9)$ → $x' = -1 - 9 = -10$, $y' = -1 + 1 = 0$ → $(-10,0)$
3. $(-1,5)$ → $x' = -1 - 5 = -6$, $y' = -1 + 1 = 0$ → $(-6,0)$
4. $(-3,3)$ → $x' = -1 - 3 = -4$, $y' = -3 + 1 = -2$ → $(-4,-2)$

Final Shape L: $(-8,-2), (-10,0), (-6,0), (-4,-2)$

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✔ Summary of Final Shapes:

| Problem | Transformation Steps | Final Coordinates |
|--------|-----------------------|-------------------|
| 1 | Reflect in y-axis → translate $\begin{pmatrix} -2 \\ -6 \end{pmatrix}$ | $(3,-1), (1,-1), (1,-2), (3,-2), (3,-3), (1,-3), (1,-4), (3,-4)$ |
| 2 | Rotate 180° about (-1,2) → reflect in $y=x$ | $(1,-2), (1,-6), (4,-2)$ |
| 3 | Enlarge sf -2 from (2,-3) → translate $\begin{pmatrix} -4 \\ 3 \end{pmatrix}$ | $(-2,4), (-4,4), (-4,0), (-2,0), (-2,-4), (0,-4), (0,0)$ |
| 4 | Rotate 90° CW about origin → reflect in $x=-3$ | $(-3,4), (-3,2), (-1,2), (-1,4)$ |
| 5 | Enlarge sf ½ from (0,-1) → reflect in $y=-x$ | $(-2,1), (-2,-2), (0,-2)$ |
| 6 | Translate $\begin{pmatrix} 0 \\ 4 \end{pmatrix}$ → rotate 90° ACW about (-1,0) | $(-8,-2), (-10,0), (-6,0), (-4,-2)$ |

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📌 How to Draw:

For each problem:
1. Identify original shape vertices.
2. Apply transformations in order.
3. Plot the new points on the grid.
4. Connect them to form the transformed shape.

These steps will help you complete all transformations accurately.

Let me know if you'd like a visual sketch or printable version!