Topic 7: Transformations and vectors

Transformations move or resize shapes on the coordinate grid, and vectors are the language used to describe much of that movement. In this topic you learn to perform and recognise the four key transformations, to describe any single transformation completely (which is where most marks are lost or won), and to combine two transformations into one. Vectors then extend the same idea of directed movement into algebra: you add them, subtract them, scale them, measure their length, and use them to prove geometric facts. Mastering both halves together gives you a powerful, consistent way to think about position and movement in the plane.

Translations and column vectors

A translation slides every point of a shape the same distance in the same direction, with no turning and no resizing. The original shape is the object and the result is the image; the two are always congruent (identical in size and shape).\n\nA translation is described by a column vector $\\binom{x}{y}$, where the top number is movement in the x-direction (right is positive, left is negative) and the bottom number is movement in the y-direction (up is positive, down is negative). For example, $\\binom{3}{-2}$ means move 3 units right and 2 units down.\n\nTo translate a point, add the column vector to its coordinates. The point $(1, 4)$ under $\\binom{3}{-2}$ becomes $(1+3, 4-2) = (4, 2)$. Apply this to every vertex of a shape and join the new points.\n\nTo find the vector that maps one shape onto another, subtract: take an image vertex and subtract the matching object vertex. If $A = (2, 5)$ maps to $A' = (6, 3)$, the translation vector is $\\binom{6-2}{3-5} = \\binom{4}{-2}$. The reverse (undoing) translation is simply the negative, $\\binom{-4}{2}$.

Reflections

A reflection flips a shape across a mirror line, producing a congruent image that is laterally inverted (back-to-front). Every point and its image lie on opposite sides of the mirror line, the same perpendicular distance away, so the mirror line is the perpendicular bisector of the segment joining a point to its image.\n\nThe common mirror lines in IGCSE are the axes and simple diagonals. Reflecting in the x-axis (the line $y = 0$) sends $(x, y)$ to $(x, -y)$. Reflecting in the y-axis (the line $x = 0$) sends $(x, y)$ to $(-x, y)$. Reflecting in the line $y = x$ swaps the coordinates to $(y, x)$, and reflecting in $y = -x$ gives $(-y, -x)$. You can also be asked to reflect in vertical lines such as $x = 2$ or horizontal lines such as $y = -1$.\n\nA reliable method: from each vertex, measure the perpendicular distance to the mirror line, then step the same distance to the other side. Points that lie on the mirror line do not move. To fully describe a reflection you must state that it is a reflection and give the equation of the mirror line.

Rotations

A rotation turns a shape about a fixed point called the centre of rotation, through a given angle and direction. The image is congruent to the object, and every point stays the same distance from the centre as it turns.\n\nThree pieces of information define a rotation completely: the angle (commonly 90, 180 or 270 degrees), the direction (clockwise or anticlockwise; note that a 180 degree rotation needs no direction since both give the same result), and the centre of rotation.\n\nFor a rotation about the origin, the coordinate rules are worth knowing. A 90 degree anticlockwise turn sends $(x, y)$ to $(-y, x)$. A 90 degree clockwise turn sends $(x, y)$ to $(y, -x)$. A 180 degree turn sends $(x, y)$ to $(-x, -y)$.\n\nWhen the centre is not the origin, tracing paper is the most dependable exam tool: place it over the object, push a pencil point onto the centre, and rotate. To find an unknown centre, join two object points to their images and construct the perpendicular bisector of each segment; the centre lies where the bisectors cross.

Enlargements, including negative and fractional scale factors

An enlargement changes the size of a shape by a scale factor $k$ from a fixed point called the centre of enlargement. Unlike the other three transformations, an enlargement (unless $k = 1$ or $k = -1$) is not congruent: the image is similar to the object, with all lengths multiplied by $k$ and all angles unchanged. Areas are multiplied by $k^2$.\n\nThe scale factor controls the result. If $k \\gt 1$ the image is larger; if $0 \\lt k \\lt 1$ it is a fractional enlargement and the image is smaller. A negative scale factor places the image on the opposite side of the centre and turns it upside down: for example $k = -2$ doubles the lengths and rotates the shape 180 degrees about the centre.\n\nTo construct an enlargement, draw a ray from the centre $O$ through each vertex. The image of a vertex sits on that ray (or its extension backwards for negative $k$) at $k$ times the distance from $O$. Algebraically, if the centre is $O$ and a vertex is at displacement $\\binom{x}{y}$ from $O$, the image is at displacement $k\\binom{x}{y}$. To find an unknown centre, draw rays through pairs of corresponding object and image points and extend them until they meet.

Describing and combining transformations

Examiners reward complete descriptions and penalise vague ones, so learn the required wording for each type. State a single transformation only, never two, and give exactly the details below:\n\nTranslation: the word translation and the column vector, e.g. translation by $\\binom{4}{-2}$. Reflection: the word reflection and the equation of the mirror line, e.g. reflection in $y = x$. Rotation: the word rotation, the angle, the direction (except for 180 degrees) and the centre, e.g. rotation 90 degrees anticlockwise about $(0, 0)$. Enlargement: the word enlargement, the scale factor and the centre, e.g. enlargement scale factor 3, centre $(1, 2)$.\n\nA combined transformation applies one transformation and then another to the result. Carry out the first to get an intermediate image, then apply the second to that image. Often a single equivalent transformation produces the same final image; for instance, two reflections in parallel mirror lines combine into one translation, while two reflections in perpendicular lines combine into a 180 degree rotation. When asked for the single equivalent transformation, compare the original object directly with the final image and describe it fully.

Vectors: notation, arithmetic and magnitude

A vector has both magnitude (size) and direction, and is written as a column vector $\\binom{x}{y}$ in the same way as a translation. Vectors are often named with a bold or underlined letter such as $\\mathbf{a}$, or by their endpoints with an arrow, $\\overrightarrow{AB}$, meaning the vector from $A$ to $B$.\n\nTo add vectors, add the components: $\\binom{3}{1} + \\binom{-1}{4} = \\binom{2}{5}$. Geometrically this is the nose-to-tail rule, following one vector then the next. To subtract, subtract the components: $\\binom{3}{1} - \\binom{-1}{4} = \\binom{4}{-3}$. The vector $\\overrightarrow{BA}$ is the negative of $\\overrightarrow{AB}$, so it has the same length but the opposite direction.\n\nMultiplying by a scalar stretches the vector: $3\\binom{2}{-1} = \\binom{6}{-3}$. A scalar multiple is always parallel to the original vector. If one vector is a scalar multiple of another, the two are parallel; if the points share a common point as well, they are collinear (on a straight line).\n\nThe magnitude (length) of $\\binom{x}{y}$ is found with Pythagoras: $|\\binom{x}{y}| = \\sqrt{x^2 + y^2}$. For example, $|\\binom{3}{-4}| = \\sqrt{9 + 16} = \\sqrt{25} = 5$.

Position vectors and vector geometry

A position vector gives the location of a point relative to the origin $O$. The position vector of point $A$ is $\\overrightarrow{OA} = \\mathbf{a}$, and its components are just the coordinates of $A$. Position vectors let you express any journey between points: $\\overrightarrow{AB} = \\overrightarrow{OB} - \\overrightarrow{OA} = \\mathbf{b} - \\mathbf{a}$. Read this as go back from $A$ to $O$, then out from $O$ to $B$.\n\nThis subtraction rule is the engine of most vector geometry questions. Suppose $M$ is the midpoint of $AB$; then $\\overrightarrow{OM} = \\mathbf{a} + \\frac{1}{2}(\\mathbf{b} - \\mathbf{a}) = \\frac{1}{2}(\\mathbf{a} + \\mathbf{b})$. More generally a point dividing $AB$ in a given ratio is reached by adding the right fraction of $\\overrightarrow{AB}$ to $\\mathbf{a}$.\n\nTo prove two lines are parallel, show one displacement vector is a scalar multiple of the other. To prove three points are collinear, show that two vectors joining them are scalar multiples and share a point. Always build each required vector by routing through known points and the origin, simplifying in terms of the named base vectors such as $\\mathbf{a}$ and $\\mathbf{b}$, and keep the components or letters tidy so the scalar relationship is easy to spot.