Reflection                          Rotation                         Translation               Enlargement              Stretch                                  Shear


To describe, state:                 To describe, state:              To describe, state:       To describe, state:      To describe, state:                      To describe, state:

Equation of line of reflection      Angle                            Column vector             Centre                   Invariant line                           Invariant line
                                    Direction                                                  Scale factor             Scale factor                             Scale factor
                                    Centre
-                                   -                                -
                                                                                               Scale factor             Scale factor                             Scale factor

                                                                                               = distance of A’E        = distance of A’ from invariant line     = distance of AA’

                                                                                                distance of AE           distance of A from the invariant line    distance of A from invariant

                                                                                               = Length of A’B’         *Stretch image is perpendicular to       * Sheared image moves
                                                                                                                        invariant line                           parallel to the invariant line
                                                                                                Length of AB


To draw image:                      To draw image:                   To draw image:            To draw image:           To draw image:                           To draw image:

- Drop a perpendicular from A       - Join A to centre of rotation   - move each points x      - Join A to E and        * Given stretch factor and invariant     1. Extend 2 pairs of
to                                    [R].                           units along the x axis      extend the line.       line                                     corresponding sides.
  the line of reflection. [Draw a   - Measure angle of rotation in   and y units along the y   - Mark A’ such that      -Join A to invariant line                2. Mark the intersection
  produce]                            specified direction of RA      axis according to the        EA’ = k EA            (perpendicular)                          points.
- Mark A’ on the produce where      - Use a compass to mark A’                                 - Repeat for the other   Therefore, stretch image will be         3. The line passes through the
                                                                                     x
                                                                     column vector, 
                                                                                      y .
  distance of A to the produce =      where RA = RA’                                             points                 perpendicular to the invariant           two points is the invariant
  distance of A’ to the produce.    - Repeat for other points.                                - join all points to     i.e. LA’=kLA                             line.
                                    - Join the points to complete                              form image.            Note: If k<0 (neg), then image lies on   **Note: Pick the sides that are
                                       the image.                    Note:                                              opposite sides of invariant line         not parallel. A shear is a non-
                                                                                               Note:                                                             isometric transformation that
                                                                     x > 0 – move points to                                                                      preserves the area of the
                                                                     the right                 k>1                                                               figure
                                                                     x < 0 – move points to    - increase in size
                                                                     the left                  - image and original
                                                                     y > 0 – move points       lie on the same side
                                                                     up
                                                                     y < 0 – move points       0<k<1
                                                                     down                      – decrease in size
                                                                                               - image and original
                                                                                               lie on opposite sides
To construct line of reflection:   To locate centre of rotation:                             To find centre of
                                                                                             Enlargement:
- Join A to A’                     - Join A to A’ and B to B’
- Construct a perpendicular        - The intersection is the centre                          - Draw a line through
  bisector. The perpendicular        of rotation.                                            AA’
  bisector is the line of                                                                    - Draw a line through
  reflection.                                                                                BB’
                                                                                             - The intersection of
                                                                                             AA’ and BB’ is the
                                                                                             centre of the
                                                                                             enlargement.

                                   To find the angle of rotation:                                                    To find invariant line:                To find invariant line:
                                                                                                                     1.original line’s and new line’s       1.original line’s and new
                                   - Join A to R and A’ to R                                                         intersection                           line’s intersection
                                   - Angle between the two lines                                                     2.repeat step 1                        2.repeat step 1
                                     is the angle of rotation.                                                       ***never use parallel lines            ***never use parallel lines


Area of image = area of original   Area of image = area of            Area of image = area   Area of image           Single stretch:                        Area of image is the same as
                                   original                           of original            = k² × area of                                                 the area of the object. [i.e.
                                                                                             original                Area of image = k × area of original   area remains unchanged
                                                                                             To find k:
                                                                                             Just use ratio of       Double stretch:
                                                                                             corresponding sides
                                                                                             because the triangles
                                                                                                                     Area of image =   k1 k 2   × area of
                                                                                             are similar.            original

Transformationnotes edited (1)

  • 1.
    Reflection Rotation Translation Enlargement Stretch Shear To describe, state: To describe, state: To describe, state: To describe, state: To describe, state: To describe, state: Equation of line of reflection Angle Column vector Centre Invariant line Invariant line Direction Scale factor Scale factor Scale factor Centre - - - Scale factor Scale factor Scale factor = distance of A’E = distance of A’ from invariant line = distance of AA’ distance of AE distance of A from the invariant line distance of A from invariant = Length of A’B’ *Stretch image is perpendicular to * Sheared image moves invariant line parallel to the invariant line Length of AB To draw image: To draw image: To draw image: To draw image: To draw image: To draw image: - Drop a perpendicular from A - Join A to centre of rotation - move each points x - Join A to E and * Given stretch factor and invariant 1. Extend 2 pairs of to [R]. units along the x axis extend the line. line corresponding sides. the line of reflection. [Draw a - Measure angle of rotation in and y units along the y - Mark A’ such that -Join A to invariant line 2. Mark the intersection produce] specified direction of RA axis according to the EA’ = k EA (perpendicular) points. - Mark A’ on the produce where - Use a compass to mark A’ - Repeat for the other Therefore, stretch image will be 3. The line passes through the x column vector,   y . distance of A to the produce = where RA = RA’ points perpendicular to the invariant two points is the invariant distance of A’ to the produce. - Repeat for other points.  - join all points to i.e. LA’=kLA line. - Join the points to complete   form image. Note: If k<0 (neg), then image lies on **Note: Pick the sides that are the image. Note: opposite sides of invariant line not parallel. A shear is a non- Note: isometric transformation that x > 0 – move points to preserves the area of the the right k>1 figure x < 0 – move points to - increase in size the left - image and original y > 0 – move points lie on the same side up y < 0 – move points 0<k<1 down – decrease in size - image and original lie on opposite sides
  • 2.
    To construct lineof reflection: To locate centre of rotation: To find centre of Enlargement: - Join A to A’ - Join A to A’ and B to B’ - Construct a perpendicular - The intersection is the centre - Draw a line through bisector. The perpendicular of rotation. AA’ bisector is the line of - Draw a line through reflection. BB’ - The intersection of AA’ and BB’ is the centre of the enlargement. To find the angle of rotation: To find invariant line: To find invariant line: 1.original line’s and new line’s 1.original line’s and new - Join A to R and A’ to R intersection line’s intersection - Angle between the two lines 2.repeat step 1 2.repeat step 1 is the angle of rotation. ***never use parallel lines ***never use parallel lines Area of image = area of original Area of image = area of Area of image = area Area of image Single stretch: Area of image is the same as original of original = k² × area of the area of the object. [i.e. original Area of image = k × area of original area remains unchanged To find k: Just use ratio of Double stretch: corresponding sides because the triangles Area of image = k1 k 2 × area of are similar. original