EduNinja

E7.1 Transformations

A reflection places every image point on the opposite side of the mirror line at the same perpendicular distance. The mirror line is therefore the perpendicular bisector of every point-to-image segment.

Mirror line	Point rule
$x=a$	$(x,y)\mapsto(2a-x,y)$
$y=b$	$(x,y)\mapsto(x,2b-y)$
$y=x$	$(x,y)\mapsto(y,x)$
$y=-x$	$(x,y)\mapsto(-y,-x)$
$y=x+c$	$(x,y)\mapsto(y-c,x+c)$
$y=-x+c$	$(x,y)\mapsto(c-y,c-x)$

For any other straight mirror line, draw the perpendicular from each vertex to the line, measure its distance, and mark the image the same distance beyond the line. A vertex on the mirror line stays fixed. Join image vertices in the same order with a ruler.

Reflect $P(4,-1)$ in $y=x+2$ . Using $(x,y)\mapsto(y-2,x+2)$ gives $P'=(-3,6)$ . The midpoint $\left(\frac12,\frac52\right)$ lies on $y=x+2$ , and $PP'$ is perpendicular to that line, confirming the result.

A complete description states reflection and the exact mirror line. Swapping coordinates works only for $y=x$ ; an offset diagonal line requires its offset rule or the perpendicular-distance construction.

A rotation turns every point through the same angle and direction around one fixed centre. Each point keeps its distance from the centre, so lengths, angles and area are unchanged.

Rotation about the origin	Coordinate rule
$90^\circ$ anticlockwise	$(x,y)\mapsto(-y,x)$
$90^\circ$ clockwise	$(x,y)\mapsto(y,-x)$
$180^\circ$	$(x,y)\mapsto(-x,-y)$

For centre $C=(h,k)$ , first subtract $C$ from a vertex, rotate that centre-to-point vector using the table, then add $C$ back. Repeat for every vertex and join the images in order with a ruler.

Rotate $P(7,3)$ by $90^\circ$ anticlockwise about $C=(5,1)$ . The relative vector is $(2,2)$ ; it rotates to $(-2,2)$ , so $P'=C+(-2,2)=(3,3)$ . Both $P$ and $P'$ are $\sqrt8$ units from $C$ .

A complete description gives rotation, centre, angle and clockwise or anticlockwise direction; direction need not be stated for $180^\circ$ . Do not use an origin rule on raw coordinates when the stated centre is elsewhere.

An enlargement with centre $C$ multiplies every centre-to-point vector by the same scale factor $k$ . This single rule controls the image's size, side of the centre and exact position.

P'=C+k(P-C)

Scale factor	Position of image point	Length multiplier
$k>1$	same ray, beyond the original	$k$
$0<k<1$	same ray, between centre and original	$k$
$k<0$	opposite ray from the centre	$\|k\|$

Let $C=(1,0)$ , $P=(2,1)$ and $k=-2$ . Since $P-C=(1,1)$ , multiplying gives $(-2,-2)$ and $P'=C+(-2,-2)=(-1,-2)$ . The image is opposite $P$ across $C$ and twice as far away.

A complete description gives enlargement, centre and signed scale factor. Lengths multiply by $|k|$ and areas by $k^2$ ; multiplying the raw coordinates by $k$ is valid only when the centre is the origin. Corresponding point lines meet at the centre.

A translation adds the same directed movement to every point. In a combination, perform each named transformation on the current image in the stated order; the output of one step is the input to the next.

\begin{pmatrix}x\y\end{pmatrix}\mapsto\begin{pmatrix}x+a\y+b\end{pmatrix}\quad\text{for translation vector }\begin{pmatrix}a\b\end{pmatrix}

Read the top component as horizontal movement and the bottom as vertical movement. Label an intermediate image after every transformation, apply the next rule to all of its vertices, and join straight edges with a ruler. Keeping intermediate coordinates prevents the original shape being transformed twice.

Start with $P=(2,1)$ . Reflect it in $y=x$ to obtain $P_1=(1,2)$ , then translate by $\begin{pmatrix}-3\\4\end{pmatrix}$ to obtain $P_2=(-2,6)$ . Reversing the order gives a different result, so the written sequence matters.

A vector records movement, not destination coordinates. Translation alone preserves size, shape and orientation. A combination must be described step by step: there is generally no single equivalent named transformation, and changing the order can change the final image.

E7.2 Vectors in two dimensions

A two-dimensional vector records a directed movement: its top component is the horizontal change and its bottom component is the vertical change. The same vector can be written as a column vector, a directed segment such as $\overrightarrow{AB}$ , or a bold letter such as $\mathbf a$ .

\overrightarrow{AB}=B-A=\begin{pmatrix}x_B-x_A\y_B-y_A\end{pmatrix}

Component	Positive	Negative
top	right	left
bottom	up	down

For $A=(6,4)$ and $B=(2,7)$ , $\overrightarrow{AB}=\begin{pmatrix}2-6\\7-4\end{pmatrix}=\begin{pmatrix}-4\\3\end{pmatrix}$ . This means 4 units left and 3 units up. Conversely, if $A=(4,1)$ and $\overrightarrow{AB}=\begin{pmatrix}-3\\1\end{pmatrix}$ , then $B=(1,2)$ .

Always subtract end minus start. Reversing the direction reverses both signs: $\overrightarrow{BA}=-\overrightarrow{AB}$ . This objective describes direction and components; calculating vector magnitude belongs to E7.3.

Vectors add by combining their horizontal changes and their vertical changes separately. Geometrically, addition joins movements head-to-tail, so a route through an intermediate point has the same resultant as the direct route.

\begin{pmatrix}a\b\end{pmatrix}+\begin{pmatrix}c\d\end{pmatrix}=\begin{pmatrix}a+c\b+d\end{pmatrix},\qquad\begin{pmatrix}a\b\end{pmatrix}-\begin{pmatrix}c\d\end{pmatrix}=\begin{pmatrix}a-c\b-d\end{pmatrix}

For points $A,B,C$ , the head-to-tail route law is $\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}$ . Subtracting a vector means adding its reverse: $\mathbf a-\mathbf b=\mathbf a+(-\mathbf b)$ .

Let $\mathbf a=\begin{pmatrix}3\\-2\end{pmatrix}$ and $\mathbf b=\begin{pmatrix}-5\\4\end{pmatrix}$ . Then $\mathbf a+\mathbf b=\begin{pmatrix}-2\\2\end{pmatrix}$ , while $\mathbf a-\mathbf b=\begin{pmatrix}8\\-6\end{pmatrix}$ . The different second result comes from subtracting both components of $\mathbf b$ .

Keep corresponding components aligned; do not add the top of one vector to the bottom of another. Route order matters when directed segments are named, because $\overrightarrow{AB}$ and $\overrightarrow{BA}$ have opposite signs.

Multiplying a vector by a scalar $k$ multiplies every component by the same number. The result stays parallel to the original vector: a positive scalar keeps its direction, a negative scalar reverses it, and zero gives the zero vector.

k\begin{pmatrix}a\b\end{pmatrix}=\begin{pmatrix}ka\kb\end{pmatrix}

Scalar $k$	Direction relative to original	Movement scale
$k>0$	same	multiplied by $k$
$k<0$	opposite	multiplied by $\|k\|$
$k=0$	no direction	zero movement

If $\mathbf v=\begin{pmatrix}-3\\7\end{pmatrix}$ , then $2\mathbf v=\begin{pmatrix}-6\\14\end{pmatrix}$ and $-\tfrac12\mathbf v=\begin{pmatrix}\tfrac32\\-\tfrac72\end{pmatrix}$ . Both components change by the same factor, so the vectors remain parallel.

Apply the scalar to both components. Adding $k$ to each component is not scalar multiplication. Notation such as $3\overrightarrow{AB}$ means three times the directed vector from $A$ to $B$ ; it does not mean a new point named $3A$ or $3B$ .

E7.3 Magnitude of a vector

The magnitude of a vector is its non-negative length. For a two-dimensional vector, the horizontal and vertical components form the perpendicular legs of a right triangle, so Pythagoras gives the length.

\left|\begin{pmatrix}x\y\end{pmatrix}\right|=\sqrt{x^2+y^2}

The magnitude of $\mathbf a$ is written $|\mathbf a|$ , and the magnitude of $\overrightarrow{AB}$ is written $|\overrightarrow{AB}|$ or $AB$ . Direction changes the vector but not its magnitude, so $|\overrightarrow{AB}|=|\overrightarrow{BA}|$ .

For $\mathbf v=\begin{pmatrix}-4\\5\end{pmatrix}$ , $|\mathbf v|=\sqrt{(-4)^2+5^2}=\sqrt{41}\approx6.40$ . Keep $\sqrt{41}$ when an exact value is required; otherwise round only the final value to the requested accuracy.

The same relationship can find an unknown. If $\overrightarrow{MT}=\begin{pmatrix}2k\\-k\end{pmatrix}$ and $|\overrightarrow{MT}|=\sqrt{180}$ , then $\sqrt{5k^2}=\sqrt{180}$ , so $|k|=6$ ; if $k$ is stated positive, $k=6$ .

Square each signed component before adding: $(-4)^2=16$ , not $-16$ . Magnitude is a scalar, never negative, and it has no direction. Do not add the components first or omit the square root.

E7.4 Vector geometry

A directed line segment represents both a movement and its direction. In $\overrightarrow{AB}$ , $A$ is the tail and $B$ is the head; the arrow points from the starting point to the ending point.

\overrightarrow{AB}=\begin{pmatrix}x_B-x_A\y_B-y_A\end{pmatrix}

Two directed segments represent the same vector when they have the same horizontal and vertical changes, even if they start in different places. Reversing an arrow negates the vector: $\overrightarrow{BA}=-\overrightarrow{AB}$ . Segments can be chained head-to-tail, so $\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}$ .

If $A=(-2,1)$ and $B=(3,4)$ , then $\overrightarrow{AB}=\begin{pmatrix}5\\3\end{pmatrix}$ . The segment from $C=(1,-4)$ to $D=(6,-1)$ has the same change $\begin{pmatrix}5\\3\end{pmatrix}$ , so it represents the same vector.

$\overrightarrow{AB}$ and $\overrightarrow{BA}$ lie on the same line but are not the same vector. A plain line segment has length but no chosen direction; the arrow and endpoint order are essential.

The position vector of a point $P$ is the vector from the origin to that point, written $\overrightarrow{OP}$ . It fixes the point's location, unlike a free vector that can be drawn with the same direction and length elsewhere.

\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}

If $\overrightarrow{OA}=\mathbf a$ and $\overrightarrow{OB}=\mathbf b$ , then $\overrightarrow{AB}=\mathbf b-\mathbf a$ . To find an unknown position vector, begin at $O$ , follow any valid route to the point, and add the directed vectors on that route.

Let $M$ be the midpoint of $AB$ . Since $\overrightarrow{AM}=\tfrac12(\mathbf b-\mathbf a)$ , $\overrightarrow{OM}=\mathbf a+\tfrac12(\mathbf b-\mathbf a)=\tfrac12(\mathbf a+\mathbf b)$ . The coefficients add to 1 because $M$ lies on the line through $A$ and $B$ .

A point and its position vector are related but not identical: $P$ is the point, while $\overrightarrow{OP}$ is the directed displacement from the chosen origin. Do not write $\overrightarrow{AB}=\mathbf a-\mathbf b$ when $\mathbf a=\overrightarrow{OA}$ and $\mathbf b=\overrightarrow{OB}$ .

To express a vector in terms of two coplanar vectors, choose a complete route between its endpoints, replace each segment by a known vector or a fraction of one, then collect the two vector terms.

Use arrow direction before algebra: reverse a known segment by changing its sign; use equal opposite sides in a parallelogram; multiply a whole side by the required midpoint or division fraction; finally simplify coefficients of the same vector.

AP:PB=m:n\quad\Longrightarrow\quad\overrightarrow{AP}=\frac{m}{m+n}\overrightarrow{AB}

Suppose $\overrightarrow{OA}=\mathbf a$ , $\overrightarrow{OB}=\mathbf b$ , and $M$ divides $AB$ in the ratio $AM:MB=2:3$ . Then $\overrightarrow{AB}=\mathbf b-\mathbf a$ and $\overrightarrow{AM}=\tfrac25(\mathbf b-\mathbf a)$ . Hence $\overrightarrow{OM}=\mathbf a+\overrightarrow{AM}=\tfrac35\mathbf a+\tfrac25\mathbf b$ .

The fraction attached to a segment is its share of the whole route: for $2:3$ , $AM$ is $2/5$ of $AB$ , not $2/3$ . Check the result by using a second route; both expressions must simplify to the same vector.

Vector geometry turns a diagram claim into an algebraic comparison. First derive two relevant vectors from clear routes; then factor them to expose a scalar multiple and state exactly what that proves.

Claim	Sufficient vector evidence	Required conclusion
Lines are parallel	$\mathbf u=k\mathbf v$ , $k\ne0$	same or opposite direction
$A,B,C$ are collinear	$\overrightarrow{AB}=k\overrightarrow{AC}$	both vectors share point $A$ and are parallel
$M$ is midpoint of $AB$	$\overrightarrow{AM}=\overrightarrow{MB}$	equal directed halves on the same line
$P$ divides $AB$ in $m:n$	$\overrightarrow{AP}=\frac{m}{m+n}\overrightarrow{AB}$	state $AP:PB=m:n$

If $\overrightarrow{XY}=3\mathbf a+2\mathbf b$ and $\overrightarrow{ZY}=6\mathbf a+4\mathbf b$ , then $\overrightarrow{ZY}=2\overrightarrow{XY}$ . The vectors share endpoint $Y$ and are parallel in the same direction, so $Z,X,Y$ are collinear. Also $XY=\tfrac12 ZY$ , so $X$ is the midpoint of $ZY$ .

For ratio or similarity problems, express points as weighted position vectors and compare corresponding directed sides. A common non-zero scalar factor proves parallel directions and proportional lengths; use the diagram's shared points or side order to turn that fact into the requested ratio or similarity statement.

Equal magnitudes alone do not prove parallelism or collinearity. A proof must show a scalar-multiple relationship, preserve endpoint order, and finish with the geometric conclusion rather than stopping at the algebra.

7. Transformations and vectors

Exam analysis

Most tested topics

E7.1 Transformations

Reflect a shape in any straight line

Rotate a shape about a stated centre

Enlarge with positive, fractional or negative scale factor

Translate by a vector and combine transformations

E7.2 Vectors in two dimensions

Describe directed movement with a vector

Add and subtract vectors component by component

Multiply a vector by a scalar

E7.3 Magnitude of a vector

Calculate the magnitude of a vector

E7.4 Vector geometry

Represent a vector as a directed line segment

Use position vectors from the origin

Express a route using two coplanar vectors

Use vector multiples to prove geometric relationships