Combining Transformations Worksheet | Cazoom Maths Worksheets - Free Printable

To solve the problem, we need to carefully apply each transformation step by step for each shape. Let's go through each part in detail.

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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 coordinates \((x, y)\) to \((-x, y)\).
- The vertices of Shape A are approximately:
- \((-4, 4)\)
- \((-4, 2)\)
- \((-2, 2)\)
- \((-2, 4)\)

After reflection in the y-axis:
- \((4, 4)\)
- \((4, 2)\)
- \((2, 2)\)
- \((2, 4)\)

#### Step 2: Translate the reflected shape by the vector \(\begin{pmatrix} -2 \\ -6 \end{pmatrix}\)
- Translation by \(\begin{pmatrix} -2 \\ -6 \end{pmatrix}\) means subtracting 2 from the x-coordinates and 6 from the y-coordinates.
- New coordinates:
- \((4 - 2, 4 - 6) = (2, -2)\)
- \((4 - 2, 2 - 6) = (2, -4)\)
- \((2 - 2, 2 - 6) = (0, -4)\)
- \((2 - 2, 4 - 6) = (0, -2)\)

Thus, Shape B has vertices at \((2, -2)\), \((2, -4)\), \((0, -4)\), and \((0, -2)\).

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

#### Step 1: Rotate Triangle C \(180^\circ\) about the point \((-1, 2)\)
- Rotation by \(180^\circ\) about a point \((a, b)\) changes the coordinates \((x, y)\) to \((2a - x, 2b - y)\).
- The vertices of Triangle C are approximately:
- \((-3, 3)\)
- \((0, 1)\)
- \((4, 1)\)

After rotation about \((-1, 2)\):
- For \((-3, 3)\): \((2(-1) - (-3), 2(2) - 3) = (-2 + 3, 4 - 3) = (1, 1)\)
- For \((0, 1)\): \((2(-1) - 0, 2(2) - 1) = (-2, 3)\)
- For \((4, 1)\): \((2(-1) - 4, 2(2) - 1) = (-2 - 4, 4 - 1) = (-6, 3)\)

So, the new vertices are \((1, 1)\), \((-2, 3)\), and \((-6, 3)\).

#### Step 2: Reflect the rotated triangle in the line \(y = x\)
- Reflection in the line \(y = x\) swaps the coordinates \((x, y)\) to \((y, x)\).
- New coordinates:
- \((1, 1)\) remains \((1, 1)\)
- \((-2, 3)\) becomes \((3, -2)\)
- \((-6, 3)\) becomes \((3, -6)\)

Thus, Triangle D has vertices at \((1, 1)\), \((3, -2)\), and \((3, -6)\).

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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 Shape E by a scale factor of \(-2\) from \((2, -3)\)
- Enlargement by a scale factor \(k\) from a centre \((a, b)\) transforms \((x, y)\) to \((a + k(x - a), b + k(y - b))\).
- The vertices of Shape E are approximately:
- \((3, -4)\)
- \((5, -4)\)
- \((5, -6)\)
- \((3, -6)\)

After enlargement by \(-2\) from \((2, -3)\):
- For \((3, -4)\): \((2 + (-2)(3 - 2), -3 + (-2)(-4 + 3)) = (2 - 2, -3 + 2) = (0, -1)\)
- For \((5, -4)\): \((2 + (-2)(5 - 2), -3 + (-2)(-4 + 3)) = (2 - 6, -3 + 2) = (-4, -1)\)
- For \((5, -6)\): \((2 + (-2)(5 - 2), -3 + (-2)(-6 + 3)) = (2 - 6, -3 + 6) = (-4, 3)\)
- For \((3, -6)\): \((2 + (-2)(3 - 2), -3 + (-2)(-6 + 3)) = (2 - 2, -3 + 6) = (0, 3)\)

So, the new vertices are \((0, -1)\), \((-4, -1)\), \((-4, 3)\), and \((0, 3)\).

#### Step 2: Translate the enlarged shape by the vector \(\begin{pmatrix} -4 \\ 3 \end{pmatrix}\)
- Translation by \(\begin{pmatrix} -4 \\ 3 \end{pmatrix}\) means subtracting 4 from the x-coordinates and adding 3 to the y-coordinates.
- New coordinates:
- \((0 - 4, -1 + 3) = (-4, 2)\)
- \((-4 - 4, -1 + 3) = (-8, 2)\)
- \((-4 - 4, 3 + 3) = (-8, 6)\)
- \((0 - 4, 3 + 3) = (-4, 6)\)

Thus, Shape F has vertices at \((-4, 2)\), \((-8, 2)\), \((-8, 6)\), and \((-4, 6)\).

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

#### Step 1: Rotate Shape G \(90^\circ\) clockwise about the origin
- Rotation by \(90^\circ\) clockwise about the origin changes the coordinates \((x, y)\) to \((y, -x)\).
- The vertices of Shape G are approximately:
- \((-4, -4)\)
- \((-2, -4)\)
- \((-2, -6)\)

After rotation:
- For \((-4, -4)\): \((-4, 4)\)
- For \((-2, -4)\): \((-4, 2)\)
- For \((-2, -6)\): \((-6, 2)\)

#### Step 2: Reflect the rotated shape in the line \(x = -3\)
- Reflection in the line \(x = -3\) changes the coordinates \((x, y)\) to \((-6 - x, y)\).
- New coordinates:
- For \((-4, 4)\): \((-6 - (-4), 4) = (-2, 4)\)
- For \((-4, 2)\): \((-6 - (-4), 2) = (-2, 2)\)
- For \((-6, 2)\): \((-6 - (-6), 2) = (0, 2)\)

Thus, Shape H has vertices at \((-2, 4)\), \((-2, 2)\), and \((0, 2)\).

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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 Triangle I by a scale factor of \(\frac{1}{2}\) from \((0, -1)\)
- Enlargement by a scale factor \(k\) from a centre \((a, b)\) transforms \((x, y)\) to \((a + k(x - a), b + k(y - b))\).
- The vertices of Triangle I are approximately:
- \((0, 5)\)
- \((4, 4)\)
- \((4, 1)\)

After enlargement by \(\frac{1}{2}\) from \((0, -1)\):
- For \((0, 5)\): \((0 + \frac{1}{2}(0 - 0), -1 + \frac{1}{2}(5 + 1)) = (0, -1 + 3) = (0, 2)\)
- For \((4, 4)\): \((0 + \frac{1}{2}(4 - 0), -1 + \frac{1}{2}(4 + 1)) = (2, -1 + 2.5) = (2, 1.5)\)
- For \((4, 1)\): \((0 + \frac{1}{2}(4 - 0), -1 + \frac{1}{2}(1 + 1)) = (2, -1 + 1) = (2, 0)\)

So, the new vertices are \((0, 2)\), \((2, 1.5)\), and \((2, 0)\).

#### Step 2: Reflect the enlarged triangle in the line \(y = -x\)
- Reflection in the line \(y = -x\) changes the coordinates \((x, y)\) to \((-y, -x)\).
- New coordinates:
- For \((0, 2)\): \((-2, 0)\)
- For \((2, 1.5)\): \((-1.5, -2)\)
- For \((2, 0)\): \((0, -2)\)

Thus, Triangle J has vertices at \((-2, 0)\), \((-1.5, -2)\), and \((0, -2)\).

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

#### Step 1: Translate Shape K by the vector \(\begin{pmatrix} 0 \\ 4 \end{pmatrix}\)
- Translation by \(\begin{pmatrix} 0 \\ 4 \end{pmatrix}\) means adding 4 to the y-coordinates.
- The vertices of Shape K are approximately:
- \((-4, 2)\)
- \((-2, 4)\)
- \((-2, 0)\)
- \((-4, 0)\)

After translation:
- \((-4, 2 + 4) = (-4, 6)\)
- \((-2, 4 + 4) = (-2, 8)\)
- \((-2, 0 + 4) = (-2, 4)\)
- \((-4, 0 + 4) = (-4, 4)\)

#### Step 2: Rotate the translated shape \(90^\circ\) anti-clockwise about the point \((-1, 0)\)
- Rotation by \(90^\circ\) anti-clockwise about a point \((a, b)\) changes the coordinates \((x, y)\) to \((a - (y - b), b + (x - a))\).
- New coordinates:
- For \((-4, 6)\): \((-1 - (6 - 0), 0 + (-4 + 1)) = (-1 - 6, -3) = (-7, -3)\)
- For \((-2, 8)\): \((-1 - (8 - 0), 0 + (-2 + 1)) = (-1 - 8, -1) = (-9, -1)\)
- For \((-2, 4)\): \((-1 - (4 - 0), 0 + (-2 + 1)) = (-1 - 4, -1) = (-5, -1)\)
- For \((-4, 4)\): \((-1 - (4 - 0), 0 + (-4 + 1)) = (-1 - 4, -3) = (-5, -3)\)

Thus, Shape L has vertices at \((-7, -3)\), \((-9, -1)\), \((-5, -1)\), and \((-5, -3)\).

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Final Answer:

\[
\boxed{
\begin{array}{l}
\text{1) Shape B: Vertices at } (2, -2), (2, -4), (0, -4), (0, -2). \\
\text{2) Triangle D: Vertices at } (1, 1), (3, -2), (3, -6). \\
\text{3) Shape F: Vertices at } (-4, 2), (-8, 2), (-8, 6), (-4, 6). \\
\text{4) Shape H: Vertices at } (-2, 4), (-2, 2), (0, 2). \\
\text{5) Triangle J: Vertices at } (-2, 0), (-1.5, -2), (0, -2). \\
\text{6) Shape L: Vertices at } (-7, -3), (-9, -1), (-5, -1), (-5, -3).
\end{array}
}
\]