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 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