Geometric Constructions

Construction.

1. Set compasses to MORE than half the segment $AB$.
2. From $A$, draw an arc above and below the segment.
3. From $B$ with the SAME compass setting, draw arcs that intersect the first ones at two points.
4. Draw a straight line through those two intersections — that's the perpendicular bisector.

Properties. The perpendicular bisector:

• Passes through the midpoint of $AB$.
• Is perpendicular to $AB$.
• Every point on it is equidistant from $A$ and $B$.

Mark scheme. Cambridge expects:

• Arcs visible from BOTH endpoints (showing equal radius).
• The perpendicular line drawn through the intersections.

Tip. Don't erase the construction arcs. They're part of the answer.

Construction.

1. From the vertex $V$ of the angle, draw an arc that crosses BOTH rays. Call those intersections $P$ and $Q$.
2. From $P$ with the same (or any chosen) compass setting, draw an arc inside the angle.
3. From $Q$ with the SAME setting, draw another arc inside the angle. Let them meet at $X$.
4. Draw the line $VX$ — that's the angle bisector.

Properties. The angle bisector:

• Cuts the angle into two EQUAL halves.
• Every point on it is equidistant from the two rays of the angle.

Mark scheme. Cambridge wants:

• Arcs from the vertex showing where the rays were cut.
• Arcs from $P$ and $Q$ meeting at $X$ — visible.
• The bisector line drawn from $V$ through $X$.

Construction (point $P$ off the line $\ell$).

1. From $P$, draw an arc that crosses $\ell$ at TWO points $A$ and $B$.
2. Construct the perpendicular bisector of $AB$ using the standard method.
3. The perpendicular bisector passes through $P$ — it's the perpendicular from $P$ to $\ell$.

Construction (point $P$ ON the line).

1. From $P$, draw arcs that cross $\ell$ on either side at $A$ and $B$ (equal distances).
2. Construct the perpendicular bisector of $AB$. It passes through $P$ and is perpendicular to $\ell$.

A locus (plural: loci) is the set of all points that satisfy a given geometrical condition.

Standard loci.

Condition	Locus
Equidistant from a fixed point $P$	Circle centred at $P$
Equidistant from two points $A, B$	Perpendicular bisector of $AB$
Equidistant from two intersecting lines	Pair of angle bisectors
At a fixed distance $r$ from a line	Two parallel lines at distance $r$
Inside a region (e.g. "within 5 cm of $A$")	A disc / region

Method for combined loci.

1. Draw each locus separately on the same diagram.
2. Identify the OVERLAP (intersection) — that's the locus satisfying ALL conditions.
3. Shade the answer region (often Cambridge wants a shaded area).

Worked. Find the locus of points within $4$ cm of $A$ AND closer to $B$ than to $C$.

• Locus 1: disc of radius $4$ cm centred at $A$.
• Locus 2: half-plane containing $B$, separated by the perpendicular bisector of $BC$.
• Final: the part of the disc lying in that half-plane.