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

Problem: Combining Transformations

The task involves performing a series of geometric transformations on given shapes and drawing the resulting shapes on coordinate grids. Below, I will solve each problem step by step.

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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 \((-x, y)\).
- Original vertices of Shape A (assuming the blue square):
- Top-left: \((-4, 5)\)
- Top-right: \((-2, 5)\)
- Bottom-left: \((-4, 3)\)
- Bottom-right: \((-2, 3)\)
- After reflection:
- Top-left: \((4, 5)\)
- Top-right: \((2, 5)\)
- Bottom-left: \((4, 3)\)
- Bottom-right: \((2, 3)\)

- Step 2: Translate the reflected shape by the vector \(\begin{pmatrix} -2 \\ -6 \end{pmatrix}\).
- Translation adds \((-2, -6)\) to each coordinate.
- New coordinates:
- Top-left: \((4 + (-2), 5 + (-6)) = (2, -1)\)
- Top-right: \((2 + (-2), 5 + (-6)) = (0, -1)\)
- Bottom-left: \((4 + (-2), 3 + (-6)) = (2, -3)\)
- Bottom-right: \((2 + (-2), 3 + (-6)) = (0, -3)\)

- Final Shape B:
- Vertices: \((2, -1)\), \((0, -1)\), \((2, -3)\), \((0, -3)\)

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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 \((h, k)\) changes \((x, y)\) to \((2h - x, 2k - y)\).
- Original vertices of Triangle C:
- Top: \((0, 3)\)
- Left: \((-3, 2)\)
- Right: \((4, 1)\)
- After rotation:
- Top: \((2(-1) - 0, 2(2) - 3) = (-2, 1)\)
- Left: \((2(-1) - (-3), 2(2) - 2) = (1, 2)\)
- Right: \((2(-1) - 4, 2(2) - 1) = (-6, 3)\)

- Step 2: Reflect the rotated triangle in the line \(y = x\).
- Reflection in the line \(y = x\) swaps the coordinates \((x, y)\) to \((y, x)\).
- After reflection:
- Top: \((1, -2)\)
- Left: \((2, 1)\)
- Right: \((3, -6)\)

- Final Triangle D:
- Vertices: \((1, -2)\), \((2, 1)\), \((3, -6)\)

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#### 3) Shape E is first enlarged by a scale factor of \(-2\) from the center 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 \((h, k)\) changes \((x, y)\) to \((h + k(x - h), k(y - k) + k)\).
- Original vertices of Shape E:
- Top-left: \((3, -4)\)
- Top-right: \((6, -4)\)
- Bottom-left: \((3, -6)\)
- Bottom-right: \((6, -6)\)
- After enlargement:
- Top-left: \((2 + (-2)(3 - 2), -3 + (-2)(-4 + 3)) = (2 - 2, -3 + 2) = (0, -1)\)
- Top-right: \((2 + (-2)(6 - 2), -3 + (-2)(-4 + 3)) = (2 - 8, -3 + 2) = (-6, -1)\)
- Bottom-left: \((2 + (-2)(3 - 2), -3 + (-2)(-6 + 3)) = (2 - 2, -3 + 6) = (0, 3)\)
- Bottom-right: \((2 + (-2)(6 - 2), -3 + (-2)(-6 + 3)) = (2 - 8, -3 + 6) = (-6, 3)\)

- Step 2: Translate the enlarged shape by the vector \(\begin{pmatrix} -4 \\ 3 \end{pmatrix}\).
- Translation adds \((-4, 3)\) to each coordinate.
- After translation:
- Top-left: \((0 + (-4), -1 + 3) = (-4, 2)\)
- Top-right: \((-6 + (-4), -1 + 3) = (-10, 2)\)
- Bottom-left: \((0 + (-4), 3 + 3) = (-4, 6)\)
- Bottom-right: \((-6 + (-4), 3 + 3) = (-10, 6)\)

- Final Shape F:
- Vertices: \((-4, 2)\), \((-10, 2)\), \((-4, 6)\), \((-10, 6)\)

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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 changes \((x, y)\) to \((y, -x)\).
- Original vertices of Shape G:
- Top-left: \((-4, -3)\)
- Top-right: \((-2, -3)\)
- Bottom-left: \((-4, -5)\)
- Bottom-right: \((-1, -5)\)
- After rotation:
- Top-left: \((-3, 4)\)
- Top-right: \((-3, 2)\)
- Bottom-left: \((-5, 4)\)
- Bottom-right: \((-5, 1)\)

- Step 2: Reflect the rotated shape in the line \(x = -3\).
- Reflection in the line \(x = -3\) changes \((x, y)\) to \((-6 - x, y)\).
- After reflection:
- Top-left: \((-6 - (-3), 4) = (-3, 4)\)
- Top-right: \((-6 - (-3), 2) = (-3, 2)\)
- Bottom-left: \((-6 - (-5), 4) = (-1, 4)\)
- Bottom-right: \((-6 - (-5), 1) = (-1, 1)\)

- Final Shape H:
- Vertices: \((-3, 4)\), \((-3, 2)\), \((-1, 4)\), \((-1, 1)\)

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#### 5) Triangle I is enlarged by a scale factor of \(\frac{1}{2}\), center 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 \((h, k)\) changes \((x, y)\) to \((h + k(x - h), k(y - k) + k)\).
- Original vertices of Triangle I:
- Top: \((1, 5)\)
- Left: \((-3, 4)\)
- Right: \((4, 1)\)
- After enlargement:
- Top: \((0 + \frac{1}{2}(1 - 0), -1 + \frac{1}{2}(5 + 1)) = (0.5, -1 + 3) = (0.5, 2)\)
- Left: \((0 + \frac{1}{2}(-3 - 0), -1 + \frac{1}{2}(4 + 1)) = (-1.5, -1 + 2.5) = (-1.5, 1.5)\)
- Right: \((0 + \frac{1}{2}(4 - 0), -1 + \frac{1}{2}(1 + 1)) = (2, -1 + 1) = (2, 0)\)

- Step 2: Reflect the enlarged triangle in the line \(y = -x\).
- Reflection in the line \(y = -x\) changes \((x, y)\) to \((-y, -x)\).
- After reflection:
- Top: \((-2, -0.5)\)
- Left: \((-1.5, 1.5)\)
- Right: \((0, -2)\)

- Final Triangle J:
- Vertices: \((-2, -0.5)\), \((-1.5, 1.5)\), \((0, -2)\)

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

- Step 1: Translate Shape K by the vector \(\begin{pmatrix} 0 \\ -4 \end{pmatrix}\).
- Translation adds \((0, -4)\) to each coordinate.
- Original vertices of Shape K:
- Top-left: \((-3, 4)\)
- Top-right: \((-2, 5)\)
- Bottom-left: \((-4, 2)\)
- Bottom-right: \((-1, 3)\)
- After translation:
- Top-left: \((-3 + 0, 4 + (-4)) = (-3, 0)\)
- Top-right: \((-2 + 0, 5 + (-4)) = (-2, 1)\)
- Bottom-left: \((-4 + 0, 2 + (-4)) = (-4, -2)\)
- Bottom-right: \((-1 + 0, 3 + (-4)) = (-1, -1)\)

- Step 2: Rotate the translated shape \(90^\circ\) counterclockwise about \((-1, 0)\).
- Rotation by \(90^\circ\) counterclockwise about \((h, k)\) changes \((x, y)\) to \((h - (y - k), k + (x - h))\).
- After rotation:
- Top-left: \((-1 - (0 - 0), 0 + (-3 + 1)) = (-1, -2)\)
- Top-right: \((-1 - (1 - 0), 0 + (-2 + 1)) = (-2, -1)\)
- Bottom-left: \((-1 - (-2 - 0), 0 + (-4 + 1)) = (1, -3)\)
- Bottom-right: \((-1 - (-1 - 0), 0 + (-1 + 1)) = (0, 0)\)

- Final Shape L:
- Vertices: \((-1, -2)\), \((-2, -1)\), \((1, -3)\), \((0, 0)\)

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Final Answers:

1. Shape B: \(\boxed{(2, -1), (0, -1), (2, -3), (0, -3)}\)
2. Triangle D: \(\boxed{(1, -2), (2, 1), (3, -6)}\)
3. Shape F: \(\boxed{(-4, 2), (-10, 2), (-4, 6), (-10, 6)}\)
4. Shape H: \(\boxed{(-3, 4), (-3, 2), (-1, 4), (-1, 1)}\)
5. Triangle J: \(\boxed{(-2, -0.5), (-1.5, 1.5), (0, -2)}\)
6. Shape L: \(\boxed{(-1, -2), (-2, -1), (1, -3), (0, 0)}\)