Prove geometric results using vectors, including properties of triangles, parallelograms and the diagonals of quadrilaterals, by expressing points as position vectors

What this dot point is asking

NESA wants you to prove geometric theorems using vectors rather than coordinate or Euclidean methods. You represent each point by a position vector, express segments as differences of position vectors, and translate geometric facts (parallel, equal length, perpendicular, midpoint) into vector statements. The proof then follows from vector algebra and the scalar product.

Position vectors and segments

Fix an origin $O$. Every point $P$ then corresponds to a position vector $\mathbf{p} = \overrightarrow{OP}$. The directed segment from $A$ to $B$ is

This single identity is the engine of vector geometry. Two segments $\overrightarrow{AB}$ and $\overrightarrow{CD}$ are parallel exactly when $\mathbf{b} - \mathbf{a} = \lambda(\mathbf{d} - \mathbf{c})$ for some scalar $\lambda$. They are equal in length when $|\mathbf{b} - \mathbf{a}| = |\mathbf{d} - \mathbf{c}|$, and perpendicular when $(\mathbf{b} - \mathbf{a}) \cdot (\mathbf{d} - \mathbf{c}) = 0$.

The midpoint and section formulas

The midpoint $M$ of segment $AB$ has position vector

More generally, the point $P$ dividing $AB$ internally in the ratio $m:n$ (so $AP:PB = m:n$) has position vector

This follows because $\overrightarrow{AP} = \frac{m}{m+n}\overrightarrow{AB}$, so $\mathbf{p} = \mathbf{a} + \frac{m}{m+n}(\mathbf{b} - \mathbf{a})$, which simplifies to the stated form.

Translating geometry into vectors

The strategy is consistent. Assign position vectors to the named points, often choosing the origin cleverly to simplify. Express the geometric claim as a vector equation. For a length-based claim, square the magnitude using $|\mathbf{v}|^2 = \mathbf{v} \cdot \mathbf{v}$, since the scalar product turns lengths and angles into algebra. For a parallelism claim, exhibit the scalar multiple. For a perpendicularity claim, show the scalar product vanishes.

A typical result: the diagonals of a parallelogram bisect each other. Let the parallelogram be $OABC$ with $\overrightarrow{OA} = \mathbf{a}$ and $\overrightarrow{OC} = \mathbf{c}$. Then $B$ has position vector $\mathbf{a} + \mathbf{c}$. The midpoint of diagonal $OB$ is $\frac{1}{2}(\mathbf{a} + \mathbf{c})$, and the midpoint of diagonal $AC$ is $\frac{1}{2}(\mathbf{a} + \mathbf{c})$. The two midpoints coincide, so the diagonals bisect each other.

Using the scalar product for angles and lengths

When perpendicularity or equal lengths appear, the scalar product is decisive. The key identities are

These let you prove, for example, that a triangle inscribed in a semicircle has a right angle at the circumference, by placing the centre at the origin and showing two radii-based vectors have zero scalar product.