This note covers the four transformations of a shape on a coordinate grid: reflection, rotation, translation and enlargement. You will learn what information is needed to describe each one fully, and how to find the image of a shape. Transformations are used in computer graphics, design and architecture.
Key ideas
A transformation moves or changes a shape. The original shape is the object and the new shape is the image. In a reflection, a rotation and a translation the image is the same size and shape as the object (they are congruent). In an enlargement the image has the same shape but a different size (it is similar).
When you are asked to describe a transformation, you must name the single transformation and give all the details it needs. Describing two transformations at once is wrong.
- Reflection: give the mirror line, for example "reflection in the line \(x = -1\)" or "reflection in the line \(y = x\)". Each point and its image are the same distance from the mirror line, on opposite sides, and the line joining them is perpendicular to it.
- Rotation: give the centre, the angle and the direction (clockwise or anticlockwise). A half-turn (\(180^\circ\)) needs no direction.
- Translation: give the column vector \(\begin{pmatrix} x \\ y \end{pmatrix}\). The top number is the move right (negative means left) and the bottom number is the move up (negative means down). Every point moves the same way, and the shape does not turn.
- Enlargement: give the centre and the scale factor. Every length is multiplied by the scale factor, and angles stay the same.
Enlargement details. A scale factor greater than 1 makes the shape bigger. A scale factor between 0 and 1 (a fraction) makes it smaller. The image of a point \(P\) is found by measuring from the centre \(C\): the vector from the centre to the image is the scale factor times the vector from the centre to \(P\). Core courses use positive scale factors. At Extended level you may also meet a negative scale factor: the image is on the opposite side of the centre from the object, and is also upside down (turned through \(180^\circ\)).
Finding the centre of an enlargement. Draw a straight line through each pair of matching points (object corner and image corner). The lines meet at the centre. The scale factor is a length on the image divided by the matching length on the object.
Finding a rotation. The angle can be seen by comparing a line on the object with the same line on the image. To find the centre, join two matching points, draw the perpendicular bisector of that line, then do the same for another pair; the bisectors meet at the centre. Trial and error with tracing paper also works.
Useful results for rotations about the origin. A \(90^\circ\) clockwise turn sends \((x, y)\) to \((y, -x)\). A \(90^\circ\) anticlockwise turn sends \((x, y)\) to \((-y, x)\). A half-turn sends \((x, y)\) to \((-x, -y)\).
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| Object and image | The shape before and after the transformation. |
| Congruent | Exactly the same size and shape (reflection, rotation, translation). |
| Similar | Same shape, different size (enlargement). |
| Mirror line | The line of reflection. It is the perpendicular bisector of each point and its image. |
| \(\begin{pmatrix} a \\ b \end{pmatrix}\) | Translation: move \(a\) right and \(b\) up. |
| Scale factor \(k\) | \(k = \dfrac{\text{image length}}{\text{object length}}\). The image of a point is found by multiplying its position from the centre by \(k\). |
| Half-turn about the origin | \((x, y) \to (-x, -y)\). |
Worked examples
Example 1 Triangle A has vertices \((2, 1)\), \((5, 1)\) and \((5, 3)\). Its image B has vertices \((-4, 1)\), \((-7, 1)\) and \((-7, 3)\). Describe fully the single transformation from A to B.
The shape has the same size and is not turned: the \(y\)-coordinates are unchanged, but the \(x\)-coordinates have moved to the other side of the line \(x = 0\). This suggests a reflection in a vertical line.
Take the point \((2, 1)\) and its image \((-4, 1)\). The mirror line is halfway between them: \(\dfrac{2 + (-4)}{2} = -1\).
Check with a second point: \((5, 1)\) and \((-7, 1)\) have midpoint \(\dfrac{5 + (-7)}{2} = -1\). This agrees.
Answer: reflection in the line \(x = -1\).
Example 2 Triangle A has vertices \((2, 1)\), \((4, 1)\) and \((4, 2)\). Triangle C has vertices \((1, -2)\), \((1, -4)\) and \((2, -4)\). Describe fully the single transformation from A to C.
The shape is the same size but has turned, so it is a rotation. The side from \((2, 1)\) to \((4, 1)\) is horizontal in A. The matching side from \((1, -2)\) to \((1, -4)\) is vertical in C and points downwards, so the turn is a quarter-turn (\(90^\circ\)) clockwise.
Test the rule \((x, y) \to (y, -x)\) for a \(90^\circ\) clockwise turn about the origin: \((2, 1) \to (1, -2)\), \((4, 1) \to (1, -4)\) and \((4, 2) \to (2, -4)\). All three match.
Answer: rotation of \(90^\circ\) clockwise about the origin \((0, 0)\).
Example 3 Triangle A has vertices \((2, 1)\), \((4, 1)\) and \((4, 2)\). Triangle D has vertices \((4, 3)\), \((10, 3)\) and \((10, 6)\). Describe fully the single transformation from A to D.
The shape is bigger but has the same orientation, so it is an enlargement.
Scale factor: the base of A has length \(4 - 2 = 2\) and the base of D has length \(10 - 4 = 6\). So \(k = \dfrac{6}{2} = 3\).
Centre: the line through \((2, 1)\) and \((4, 3)\) is \(y = x - 1\). The line through \((4, 1)\) and \((10, 3)\) has gradient \(\dfrac{2}{6} = \dfrac13\), so \(y - 1 = \dfrac13 (x - 4)\). Setting \(x - 1 = 1 + \dfrac{x - 4}{3}\) gives \(3x - 3 = 3 + x - 4\), so \(2x = 2\) and \(x = 1\), then \(y = 0\). The centre is \((1, 0)\).
Check: from the centre \((1, 0)\), the vertex \((2, 1)\) is \((1, 1)\) away. Multiplying by 3 gives \((3, 3)\), so the image is \((1 + 3, 0 + 3) = (4, 3)\). This matches.
Answer: enlargement, scale factor 3, centre \((1, 0)\).
Common errors
- Common error: describing a transformation with only part of the information, such as "rotation of \(90^\circ\)" with no centre or direction, or "enlargement by 3" with no centre.
- Common error: using two words for one move, such as "reflect and translate". Each description must be a single transformation.
- Common error: giving a translation as a coordinate pair or in words only. Use a column vector, with right and up as positive.
- Common error: measuring the scale factor from the wrong lengths, or the wrong way round. For a reduction, the scale factor is a fraction less than 1.
- Common error: confusing \(x = 2\) (a vertical line) with \(y = 2\) (a horizontal line) when naming a mirror line.
Exam tip: Check your answer on at least two vertices of the shape. If the transformation works for one point but not another, you have the centre, line or direction wrong.
Check yourself
- Describe the translation that moves the point \((2, 1)\) to \((-1, 5)\).
Answer
Change in \(x\): \(-1 - 2 = -3\). Change in \(y\): \(5 - 1 = 4\). The translation is \(\begin{pmatrix} -3 \\ 4 \end{pmatrix}\). - Enlarge the point \((8, 6)\) with scale factor \(\tfrac12\) and centre the origin. Where does it go?
Answer
Multiply each coordinate by \(\tfrac12\): the image is \((4, 3)\). - (Extended) The point \((3, 2)\) is enlarged with scale factor \(-2\) and centre \((1, 1)\). Find the image.
Answer
The vector from the centre to the point is \((3 - 1, 2 - 1) = (2, 1)\). Multiply by \(-2\) to get \((-4, -2)\). Add the centre: \((1 - 4, 1 - 2) = (-3, -1)\). - Find the image of the point \((3, 2)\) after a half-turn about the origin.
Answer
A half-turn about the origin changes the sign of both coordinates, so the image is \((-3, -2)\).