Position vectors and distanceEdexcel A-Level Maths: Revision notes

The position vector of a point $A$ is the vector $\overrightarrow{OA}$ from a fixed origin $O$ to $A$. It is usually written $\mathbf{a}$. The point $A(3,-2)$ has position vector $\mathbf{a}=3\mathbf{i}-2\mathbf{j}$, so the coordinates of a point are the components of its position vector. A position vector is tied to the origin, whereas a general vector, such as a displacement, can be drawn anywhere. Using position vectors lets you describe points and shapes with algebra.

To get from $A$ to $B$ you can go via the origin: back along $A$ to $O$, then out to $B$. So $\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=\mathbf{b}-\mathbf{a}.$ Remember it as 'end minus start'. For $\mathbf{a}=2\mathbf{i}+7\mathbf{j}$ and $\mathbf{b}=8\mathbf{i}-\mathbf{j}$: $\overrightarrow{AB}=6\mathbf{i}-8\mathbf{j}$, and $\overrightarrow{BA}=-\overrightarrow{AB}=-6\mathbf{i}+8\mathbf{j}$. The position vector of the midpoint $M$ of $AB$ is $\overrightarrow{OM}=\mathbf{a}+\frac12\overrightarrow{AB}=\frac12(\mathbf{a}+\mathbf{b})$, here $5\mathbf{i}+3\mathbf{j}$.

The distance between $A(x_1,y_1)$ and $B(x_2,y_2)$ is the magnitude of $\overrightarrow{AB}$, by Pythagoras: $d^2=(x_1-x_2)^2+(y_1-y_2)^2,\qquad AB=|\mathbf{b}-\mathbf{a}|.$ The order of subtraction does not matter, because each difference is squared. Example: $A(2,7)$ and $B(8,-1)$ give $AB^2=6^2+(-8)^2=100$, so $AB=10$. Leave answers as surds when an exact value is asked for, such as $\sqrt{40}=2\sqrt{10}$.

Three points $P$, $Q$, $R$ are collinear (lie on one straight line) when $\overrightarrow{PQ}$ and $\overrightarrow{QR}$ are parallel, because they share the point $Q$. Example: $\overrightarrow{PQ}=4\mathbf{i}+3\mathbf{j}$ and $\overrightarrow{QR}=8\mathbf{i}+6\mathbf{j}=2\overrightarrow{PQ}$, so $P$, $Q$, $R$ are collinear and $PQ:QR=1:2$. A triangle is isosceles if two sides have equal length. For $A(1,2)$, $B(7,4)$, $C(3,8)$: $AB^2=40$ and $AC^2=40$, so $AB=AC$. The line from $A$ to the midpoint of $BC$ is then perpendicular to $BC$ and is the height, so the area is $\frac12\times BC\times AM=16$.

A triangle has a right angle when the squared lengths satisfy Pythagoras, $AB^2+AC^2=BC^2$ (the converse of Pythagoras). For $A(2,1)$, $B(8,5)$, $C(0,4)$: $52+13=65$, so angle $A=90^\circ$ and the area is $\frac12\sqrt{52}\sqrt{13}=13$. In a rectangle or parallelogram $ABDC$, opposite sides are equal vectors: $\overrightarrow{BD}=\overrightarrow{AC}$. So $\mathbf{d}=\mathbf{b}+\overrightarrow{AC}=\mathbf{b}+\mathbf{c}-\mathbf{a}$, here $6\mathbf{i}+8\mathbf{j}$. The diagonals bisect each other: the midpoints of $AD$ and $BC$ are the same point.