Mastering Transformations: A Guide to Reflection, Rotation, Translation

What the examiner is testing

The examiner is assessing your ability to accurately apply a given transformation (reflection, rotation, translation, or enlargement) to a 2D shape on a coordinate grid, or to describe a transformation that maps one shape onto another. Marks are awarded for correct positioning of the transformed shape and for providing all necessary descriptive information (e.g., centre of rotation, scale factor).

The method

For reflecting a shape

1. Draw the mirror line, if it's not already drawn.
2. For each vertex of the original shape, measure the perpendicular distance from the vertex to the mirror line.
3. Measure the same distance on the opposite side of the mirror line, along the perpendicular path, and mark the new vertex.
4. Once all vertices are reflected, connect them in the correct order to form the image.

For rotating a shape

1. Place your tracing paper over the original shape and the coordinate grid.
2. Trace the original shape and mark the centre of rotation on the tracing paper.
3. Rotate the tracing paper around the centre of rotation by the given angle and direction (clockwise or anti-clockwise).
4. Once rotated, press firmly to make impressions of the new vertices onto the grid below. Remove the tracing paper and connect these new vertices to form the image.

For translating a shape

1. Identify the translation vector \(\begin{pmatrix} x \\ y \end{pmatrix}\), where \(x\) is the horizontal movement and \(y\) is the vertical movement.
2. For each vertex of the original shape, move it \(x\) units horizontally (right for positive, left for negative) and \(y\) units vertically (up for positive, down for negative).
3. Mark the new position of each vertex.
4. Connect the new vertices in the correct order to form the image.

For enlarging a shape

1. Draw lines from the centre of enlargement through each vertex of the original shape.
2. Measure the distance from the centre of enlargement to each vertex of the original shape.
3. Multiply this distance by the scale factor.
4. Measure this new distance along the line drawn in step 1, from the centre of enlargement, and mark the new vertex. Repeat for all vertices.
5. Connect the new vertices to form the enlarged image.

For describing a transformation

1. Determine if the shape has changed orientation (reflection), rotated (rotation), moved without changing size or orientation (translation), or changed size (enlargement).
2. If it's a reflection, identify the mirror line.
3. If it's a rotation, find the centre of rotation (by trial and error with tracing paper, or by bisecting lines connecting corresponding points), the angle, and the direction.
4. If it's a translation, pick a corresponding vertex on both shapes and determine the horizontal and vertical movement to form the translation vector.
5. If it's an enlargement, find the centre of enlargement (by drawing lines through corresponding vertices and finding their intersection) and calculate the scale factor (new length / original length).

Worked example

Triangle A has vertices at \((1, 2)\), \((1, 4)\) and \((3, 2)\).
Translate Triangle A by the vector \(\begin{pmatrix} -2 \\ 3 \end{pmatrix}\). Label the image Triangle B.

• Original vertices of Triangle A:
  • \((1, 2)\)
  • \((1, 4)\)
  • \((3, 2)\)
• Translation vector: \(\begin{pmatrix} -2 \\ 3 \end{pmatrix}\)
• Apply the translation to each vertex:
  • For \((1, 2)\): \( (1 + (-2), 2 + 3) = (-1, 5) \)
  • For \((1, 4)\): \( (1 + (-2), 4 + 3) = (-1, 7) \)
  • For \((3, 2)\): \( (3 + (-2), 2 + 3) = (1, 5) \)
• Vertices of Triangle B: \((-1, 5)\), \((-1, 7)\) and \((1, 5)\).
• Plot these new vertices and connect them to form Triangle B.

Sanity check: Triangle B should be congruent to Triangle A and shifted 2 units left and 3 units up from its original position. The orientation should be identical.

Worked example: a harder one

Shape P is reflected in the line \(y = 1\) to give shape Q. Shape Q is then rotated \(90^\circ\) clockwise about the point \((1, 1)\) to give shape R.
Describe the single transformation that maps shape P to shape R.

• Initial thought: Just reflect and then rotate. This will give shape R, but the question asks for a single transformation. Combining transformations usually requires more than just listing them.

• Step 1: Reflect Shape P in the line \(y = 1\).
  Let's assume Shape P has vertices at \((2, 3), (4, 3), (4, 5)\).
  The line \(y=1\) is a horizontal line.

  • \((2, 3)\) is 2 units above \(y=1\). Reflected point: \((2, 1-2) = (2, -1)\).
  • \((4, 3)\) is 2 units above \(y=1\). Reflected point: \((4, 1-2) = (4, -1)\).
  • \((4, 5)\) is 4 units above \(y=1\). Reflected point: \((4, 1-4) = (4, -3)\).
    Shape Q has vertices at \((2, -1), (4, -1), (4, -3)\).

• Step 2: Rotate Shape Q \(90^\circ\) clockwise about \((1, 1)\).
  Using tracing paper or coordinate rules for rotation about a point \((a, b)\):
  For a \(90^\circ\) clockwise rotation about \((a, b)\), a point \((x, y)\) maps to \((a + (y-b), b - (x-a))\).
  Here, \((a, b) = (1, 1)\).

  • For \((2, -1)\) (vertex of Q):
    \( (1 + (-1-1), 1 - (2-1)) = (1 + (-2), 1 - 1) = (-1, 0) \)
  • For \((4, -1)\) (vertex of Q):
    \( (1 + (-1-1), 1 - (4-1)) = (1 + (-2), 1 - 3) = (-1, -2) \)
  • For \((4, -3)\) (vertex of Q):
    \( (1 + (-3-1), 1 - (4-1)) = (1 + (-4), 1 - 3) = (-3, -2) \)
    Shape R has vertices at \((-1, 0), (-1, -2), (-3, -2)\).

• Step 3: Describe the single transformation from P to R.
  Compare Shape P (\((2, 3), (4, 3), (4, 5)\)) with Shape R (\((-1, 0), (-1, -2), (-3, -2)\)).
  The orientation of the shape has changed, but its size has not. This suggests a rotation or reflection.
  If it were a reflection, the orientation would be flipped. Here, it appears to have rotated.
  Let's try to find the centre and angle of rotation.
  Visually, P is in the top-right quadrant, R is in the bottom-left. This suggests a \(180^\circ\) rotation.
  If it's a \(180^\circ\) rotation, the centre of rotation \((h, k)\) is the midpoint of the line segment connecting corresponding points.
  Let's take \((2, 3)\) from P and \((-1, 0)\) from R.
  Midpoint: \( \left( \frac{2 + (-1)}{2}, \frac{3 + 0}{2} \right) = \left( \frac{1}{2}, \frac{3}{2} \right) = (0.5, 1.5) \)
  Let's check with another pair: \((4, 3)\) from P and \((-1, -2)\) from R.
  Midpoint: \( \left( \frac{4 + (-1)}{2}, \frac{3 + (-2)}{2} \right) = \left( \frac{3}{2}, \frac{1}{2} \right) = (1.5, 0.5) \)
  The midpoints are different, so it's not a \(180^\circ\) rotation about a single point.

  This means the single transformation that maps P to R is a rotation.
  To find the centre of rotation for a general rotation:
  1. Draw a line connecting a vertex on P to its corresponding vertex on R.
  2. Draw the perpendicular bisector of this line.
  3. Repeat for a second pair of corresponding vertices.
  4. The intersection of the two perpendicular bisectors is the centre of rotation.

  Let's use \((2, 3)\) and \((-1, 0)\). Midpoint \((0.5, 1.5)\). Gradient of line is \( \frac{0-3}{-1-2} = \frac{-3}{-3} = 1 \).
  Perpendicular bisector gradient is \(-1\). Equation: \( y - 1.5 = -1(x - 0.5) \implies y = -x + 0.5 + 1.5 \implies y = -x + 2 \).

  Let's use \((4, 3)\) and \((-1, -2)\). Midpoint \((1.5, 0.5)\). Gradient of line is \( \frac{-2-3}{-1-4} = \frac{-5}{-5} = 1 \).
  Perpendicular bisector gradient is \(-1\). Equation: \( y - 0.5 = -1(x - 1.5) \implies y = -x + 1.5 + 0.5 \implies y = -x + 2 \).

  Both perpendicular bisectors are the same line \(y = -x + 2\). This indicates a problem with my assumed vertices or a more complex rotation. Let's re-evaluate the visual.
  The shape has been flipped and moved. This is a reflection.
  Let's reconsider the orientation.
  P: \((2, 3), (4, 3), (4, 5)\) (right angle at \((4, 3)\))
  R: \((-1, 0), (-1, -2), (-3, -2)\) (right angle at \((-1, -2)\))
  The "right angle" vertex has moved from \((4,3)\) to \((-1,-2)\).
  The side from \((2,3)\) to \((4,3)\) (horizontal) maps to the side from \((-1,0)\) to \((-1,-2)\) (vertical). This is a \(90^\circ\) rotation.

  Let's re-check the rotation of Q to R.
  Q: \((2, -1), (4, -1), (4, -3)\)
  Rotate \(90^\circ\) clockwise about \((1, 1)\).
  \((2, -1)\) becomes \((1 + (-1-1), 1 - (2-1)) = (1-2, 1-1) = (-1, 0)\). Correct.
  \((4, -1)\) becomes \((1 + (-1-1), 1 - (4-1)) = (1-2, 1-3) = (-1, -2)\). Correct.
  \((4, -3)\) becomes \((1 + (-3-1), 1 - (4-1)) = (1-4, 1-3) = (-3, -2)\). Correct.
  So R is correct.

  Now, P to R.
  P: \((2, 3), (4, 3), (4, 5)\)
  R: \((-1, 0), (-1, -2), (-3, -2)\)
  This is a rotation. The angle is \(90^\circ\) or \(270^\circ\).
  Let's try \(90^\circ\) anti-clockwise about some point \((h, k)\).
  A point \((x, y)\) rotated \(90^\circ\) anti-clockwise about \((h, k)\) maps to \((h - (y-k), k + (x-h))\).
  Let \((2, 3)\) map to \((-1, 0)\).
  \( h - (3-k) = -1 \implies h - 3 + k = -1 \implies h + k = 2 \) (Equation 1)
  \( k + (2-h) = 0 \implies k + 2 - h = 0 \implies k - h = -2 \) (Equation 2)
  Add (1) and (2): \( (h+k) + (k-h) = 2 + (-2) \implies 2k = 0 \implies k = 0 \).
  Substitute \(k=0\) into (1): \( h + 0 = 2 \implies h = 2 \).
  So the centre of rotation is \((2, 0)\).
  Let's check this with another point, say \((4, 3)\) from P.
  Rotate \((4, 3)\) \(90^\circ\) anti-clockwise about \((2, 0)\):
  \( (2 - (3-0), 0 + (4-2)) = (2 - 3, 0 + 2) = (-1, 2) \).
  This should map to \((-1, -2)\) in R. It doesn't. So it's not a \(90^\circ\) anti-clockwise rotation about \((2, 0)\).

  This means it must be a \(90^\circ\) clockwise rotation.
  A point \((x, y)\) rotated \(90^\circ\) clockwise about \((h, k)\) maps to \((h + (y-k), k - (x-h))\).
  Let \((2, 3)\) map to \((-1, 0)\).
  \( h + (3-k) = -1 \implies h + 3 - k = -1 \implies h - k = -4 \) (Equation 3)
  \( k - (2-h) = 0 \implies k - 2 + h = 0 \implies h + k = 2 \) (Equation 4)
  Add (3) and (4): \( (h-k) + (h+k) = -4 + 2 \implies 2h = -2 \implies h = -1 \).
  Substitute \(h=-1\) into (4): \( -1 + k = 2 \implies k = 3 \).
  So the centre of rotation is \((-1, 3)\).
  Let's check this with another point, say \((4, 3)\) from P.
  Rotate \((4, 3)\) \(90^\circ\) clockwise about \((-1, 3)\):
  \( (-1 + (3-3), 3 - (4-(-1))) = (-1 + 0, 3 - 5) = (-1, -2) \).
  This matches the corresponding vertex in R.
  Let's check the last vertex: \((4, 5)\) from P.
  Rotate \((4, 5)\) \(90^\circ\) clockwise about \((-1, 3)\):
  \( (-1 + (5-3), 3 - (4-(-1))) = (-1 + 2, 3 - 5) = (1, -2) \).
  This should match \((-3, -2)\) in R. It doesn't.

  The problem is that a reflection followed by a rotation is equivalent to a glide reflection, or if the reflection line passes through the centre of rotation, it can be a single rotation. Here, the line \(y=1\) does not pass through \((1,1)\).
  The question asks for a single transformation. This is a common trick. A reflection followed by a rotation is not always a single rotation or reflection. It's often a glide reflection, which is not taught at this level.
  Therefore, the examiner is likely looking for a combined transformation description or a specific type of single transformation if it exists.
  Let's re-examine the problem. The question implies a single transformation does exist.
  The shapes are congruent, so it's not an enlargement.
  The orientation has changed, so it's not a translation.
  Is it a reflection?
  P: \((2, 3), (4, 3), (4, 5)\)
  R: \((-1, 0), (-1, -2), (-3, -2)\)
  If it was a reflection, the "handedness" would be reversed.
  P has vertices in clockwise order \((2,3), (4,3), (4,5)\).
  R has vertices in clockwise order \((-1,0), (-1,-2), (-3,-2)\).
  The order is preserved, so it is a rotation. My earlier deduction that it was a rotation was correct. My calculations for the centre of rotation were incorrect.

  Let's use a graphical approach for the centre of rotation.
  Draw Shape P and Shape R on a grid.
  Connect corresponding points, e.g., \((2,3)\) to \((-1,0)\) and \((4,3)\) to \((-1,-2)\).
  Draw the perpendicular bisectors of these lines.
  Midpoint of \((2,3)\) and \((-1,0)\) is \((0.5, 1.5)\). Gradient of line is 1. Perpendicular bisector is \(y - 1.5 = -1(x - 0.5) \implies y = -x + 2\).
  Midpoint of \((4,3)\) and \((-1,-2)\) is \((1.5, 0.5)\). Gradient of line is 1. Perpendicular bisector is \(y - 0.5 = -1(x - 1.5) \implies y = -x + 2\).
  This still gives the same line \(y = -x + 2\). This means the angle of rotation is \(180^\circ\).
  If the angle of rotation is \(180^\circ\), then the centre of rotation is the midpoint of any pair of corresponding points.
  Let's re-check the clockwise order.
  P: \((2,3)\) (bottom-left), \((4,3)\) (bottom-right), \((4,5)\) (top-right).
  R: \((-1,0)\) (top-left), \((-1,-2)\) (bottom-left), \((-3,-2)\) (bottom-right).
  The orientation is indeed flipped. P is like an 'L' shape. R is like a 'backwards L' shape.
  Therefore, it is a reflection. My initial visual inspection was incorrect.

  It is a reflection.
  To find the mirror line:
  1. Connect a point on P to its corresponding point on R.
  2. Find the midpoint of this line segment.
  3. Find the gradient of this line segment.
  4. The mirror line passes through the midpoint and is perpendicular to the segment.
  Let's use \((2, 3)\) and \((-1, 0)\).
  Midpoint: \((0.5, 1.5)\). Gradient: \((0-3)/(-1-2) = -3/-3 = 1\).
  Gradient of mirror line: \(-1\).
  Equation of mirror line: \( y - 1.5 = -1(x - 0.5) \implies y = -x + 0.5 + 1.5 \implies y = -x + 2 \).

  Let's check with another pair: \((4, 3)\) and \((-1, -2)\).
  Midpoint: \((1.5, 0.5)\). Gradient: \((-2-3)/(-1-4) = -5/-5 = 1\).
  Gradient of mirror line: \(-1\).
  Equation of mirror line: \( y - 0.5 = -1(x - 1.5) \implies y = -x + 1.5 + 0.5 \implies y = -x + 2 \).

  Both pairs give the same mirror line.
  The single transformation is a reflection in the line \(y = -x + 2\).

Practice

1. Reflect the triangle with vertices \((1, 1), (3, 1), (3, 4)\) in the line \(x = 0\). State the coordinates of the reflected triangle.
2. Rotate the square with vertices \((0, 0), (2, 0), (2, 2), (0, 2)\) by \(90^\circ\) anti-clockwise about the origin \((0, 0)\). State the coordinates of the rotated square.
3. Triangle A has vertices \((1, 2), (4, 2), (4, 3)\). Triangle B has vertices \((-3, -1), (0, -1), (0, 0)\). Describe fully the single transformation that maps Triangle A onto Triangle B.
4. Shape S has vertices \((2, 1), (4, 1), (4, 2), (2, 2)\). Enlarge Shape S by a scale factor of \(2\) with the centre of enlargement at \((0, 0)\). Then, translate the enlarged shape by the vector \(\begin{pmatrix} -3 \\ 1 \end{pmatrix}\). What are the coordinates of the final shape?

Answers

1. Vertices of the reflected triangle: \((-1, 1), (-3, 1), (-3, 4)\).
2. Vertices of the rotated square: \((0, 0), (0, 2), (-2, 2), (-2, 0)\).
3. This is a translation.
   Pick a corresponding vertex, e.g., \((1, 2)\) from A and \((-3, -1)\) from B.
   Horizontal change: \(-3 - 1 = -4\).
   Vertical change: \(-1 - 2 = -3\).
   The translation vector is \(\begin{pmatrix} -4 \\ -3 \end{pmatrix}\).
   Transformation: Translation by vector \(\begin{pmatrix} -4 \\ -3 \end{pmatrix}\).

4. Enlargement:
   Centre of enlargement \((0, 0)\), scale factor \(2\).

   • \((2, 1) \to (2 \times 2, 1 \times 2) = (4, 2)\)
   • \((4, 1) \to (4 \times 2, 1 \times 2) = (8, 2)\)
   • \((4, 2) \to (4 \times 2, 2 \times 2) = (8, 4)\)
   • \((2, 2) \to (2 \times 2, 2 \times 2) = (4, 4)\)
     Enlarged shape vertices: \((4, 2), (8, 2), (8, 4), (4, 4)\).

   Translation:
   Translation vector \(\begin{pmatrix} -3 \\ 1 \end{pmatrix}\).
   * \((4, 2) \to (4 + (-3), 2 + 1) = (1, 3)\)
   * \((8, 2) \to (8 + (-3), 2 + 1) = (5, 3)\)
   * \((8, 4) \to (8 + (-3), 4 + 1) = (5, 5)\)
   * \((4, 4) \to (4 + (-3), 4 + 1) = (1, 5)\)
   Final coordinates: \((1, 3), (5, 3), (5, 5), (1, 5)\).

The three mistakes that lose marks

1. Missing the centre or direction for rotation: A rotation description requires three pieces of information: angle, direction (clockwise/anti-clockwise), and the centre of rotation.
   • Wrong answer: "Rotation \(90^\circ\)." (Missing direction and centre)
2. Incorrect mirror line for reflection: Especially with diagonal lines, students often reflect across \(y=x\) instead of \(y=-x\), or make sign errors.
   • Wrong answer: Reflecting \((2, 3)\) in \(y=x\) to get \((3, 2)\), but reflecting in \(y=-x\) to get \((-3, -2)\) instead of \((-3, -2)\) if the original point was \((-2, -3)\). For \((2,3)\) in \(y=-x\), the correct image is \((-3,-2)\).
3. Applying a negative scale factor incorrectly for enlargement: A negative scale factor means the enlarged image is on the opposite side of the centre of enlargement and inverted. Students often just apply the positive scale factor.
   • Wrong answer: Enlarging a point \((2, 1)\) by scale factor \(-2\) about \((0, 0)\) to get \((4, 2)\). The correct image is \((-4, -2)\).

30-second recap

Transformations move shapes. Reflections flip a shape over a line. Rotations turn a shape around a point. Translations slide a shape by a vector. Enlargements change a shape's size from a centre. Always provide all necessary information for a full description.