Transformations & Vectors

Reflection

A reflection turns a shape into its mirror image across a fixed line. Distances and angles are unchanged, so the image is congruent — only its orientation is reversed.

Describing a reflection

To describe a reflection fully you need just two things: the word reflection and the equation of the mirror line. The mirror line lies exactly halfway between each point and its image, meeting the join at a right angle.

Finding the mirror line

Given a shape and its image, join a point to its image and find the perpendicular bisector of that segment — that line is the mirror. One pair of points is enough to fix it; a second pair is a useful check.

Worked example: a triangle has vertices at (1, 4), (1, 2) and (3, 2). Its image has vertices at (4, 1), (2, 1) and (2, 3). Describe the single transformation fully. Step 1: the image is the same size and shape but reversed, so the transformation is a reflection. Step 2: each point’s coordinates have swapped, (1, 4) → (4, 1) — swapping x and y is reflection in y = x. Step 3: the midpoint of (1, 4) and (4, 1) is (2.5, 2.5), which lies on y = x — confirmed.

The four reflections worth memorising: in the x-axis, (x, y) → (x, −y); in the y-axis, (x, y) → (−x, y); in y = x, (x, y) → (y, x); in y = −x, (x, y) → (−y, −x).

Rotation

A rotation turns a shape about a fixed centre. Lengths and angles survive the turn, so the image is congruent — it has simply been swung to a new position.

Describing a rotation

A full description gives the centre, the angle and the direction. At exam level the angle is a multiple of 90°; for a 180° turn the direction makes no difference, so it may be omitted. If the centre is not the origin, find it as the point equidistant from each vertex and its image.

About the origin, the three rotations worth memorising: 90° anticlockwise, (x, y) → (−y, x); 90° clockwise, (x, y) → (y, −x); 180°, (x, y) → (−x, −y).

Finding the centre

Join two points to their images and draw the perpendicular bisector of each join. The centre of rotation is where those bisectors cross — the point that keeps its distance to every vertex.

Worked example: the point P(4, 1) is rotated 90° anticlockwise about the origin. Find the coordinates of its image P′. Step 1: for a 90° anticlockwise turn about O, apply (x, y) → (−y, x). Step 2: substitute x = 4, y = 1 to get (−1, 4). Step 3: OP and OP′ are both √17 long — the distance from the centre is preserved, as it must be. P′ = (−1, 4).

Enlargement

An enlargement changes a shape’s size while keeping its proportions. Angles are unchanged and every length is multiplied by the same scale factor, so object and image are similar.

Centre and scale factor

Measure from the centre of enlargement O. If a point sits at position P, its image P′ lies so that the distance OP′ is the scale factor times OP, along the same ray. The centre is the one point that stays put: OP′ = k × OP, where k > 1 makes the image larger, 0 < k < 1 makes it smaller, and k < 0 puts it on the opposite side of the centre, inverted.

Finding centre and factor

The scale factor is any image length divided by the matching object length. To locate the centre, draw a ray through each object point and its image; the rays all meet at the centre.

Worked example: a triangle has a vertex at P(1, 2). It is enlarged by scale factor 3 with centre the origin. Where does P map to? Step 1: with centre the origin, multiply each coordinate by the scale factor, (x, y) → (3x, 3y). Step 2: substitute (1, 2) to get (3 × 1, 3 × 2) = (3, 6). Step 3: P′ sits three times as far from O as P, along the same ray — the shape keeps its proportions. P′ = (3, 6).

Translation & Combinations

A translation slides every point of a shape the same distance in the same direction. Nothing turns and nothing resizes, so the image is congruent and the same way up.

Translation by a vector

The slide is captured by a column vector: the top number is the horizontal step, the bottom the vertical step, with left and down counting as negative. Adding the vector to each vertex gives the image — translation by (x⁄y) maps (a, b) to (a + x, b + y).

Combining transformations

Apply two transformations in turn and the result can often be described as one. Two reflections in parallel mirrors give a translation; two reflections in mirrors that cross give a rotation about the crossing point. The exam asks for that single equivalent, fully described.

Worked example: a shape is reflected in the y-axis, then the image is reflected in the line x = 3. Describe the single transformation with the same effect. Step 1: the two mirror lines, x = 0 and x = 3, are parallel, so the combined effect is a translation. Step 2: the shift is twice the gap between the mirrors, 2 × 3 = 6 units, toward the second mirror (to the right). Step 3: parallel vertical mirrors give a horizontal slide, so the vertical component is zero. Translation by (6⁄0).

Vector Arithmetic & Magnitude

A vector carries two things at once — how far and which way. In column form, vectors add, subtract and scale one component at a time.

Arithmetic in column form

Add or subtract vectors by combining the top numbers and the bottom numbers separately: (a⁄b) + (c⁄d) = (a+c⁄b+d). A scalar multiplies both components, giving a parallel vector: k(a⁄b) = (ka⁄kb). Geometrically, a + b joins tip-to-tail — the triangle law.

Magnitude of a vector

The magnitude |a| is the length of the vector, found by Pythagoras on its components: for a = (x⁄y), |a| = √(x² + y²) — always a non-negative number, not a vector.

Worked example: given a = (3⁄4) and b = (−1⁄2), find 2a − b and its magnitude. Step 1: 2a = (6⁄8). Step 2: 2a − b = (6−(−1)⁄8−2) = (7⁄6). Step 3: |2a − b| = √(7² + 6²) = √85 ≈ 9.22.

Position Vectors

Position vectors pin points to the origin, so a whole diagram can be written in terms of a few vectors. Geometric facts — parallel, collinear, a ratio along a line — then follow from algebra.

From position vectors to displacement

If A and B have position vectors a and b, the journey from A to B is the difference: go back to O, then out to B, giving AB = b − a. Expressing every segment this way lets you combine them like ordinary vectors. The midpoint M of AB has OM = ½(a + b).

Proving a geometric result

Two vectors are parallel when one is a scalar multiple of the other. If they are parallel and pass through a common point, the three endpoints are collinear. Always finish with the conclusion in words.

Worked example: in triangle OAB, OA = a and OB = b. M is the midpoint of AB. Express OM in terms of a and b. Step 1: AB = b − a, so from A the midpoint is half of this, AM = ½(b − a). Step 2: OM = OA + AM = a + ½(b − a). Step 3: tidy to a + ½b − ½a = ½a + ½b — the symmetric average of the two position vectors, as expected.