Where Should a Water Tank Stand to Be Equally Far From Two Villages?

A locus is the set of all points that satisfy a given condition, and only those points. You can also picture it as the path traced by a point that moves while obeying the condition.

For example, every point that is exactly $3$ cm from a fixed point $O$ lies on a circle of radius $3$ cm with centre $O$ — and no other point does.

This chapter covers standard loci, constructions and two-loci problems.

Imagine every position the point can take under the condition, then name the shape those positions form, including its size and position.

Examples:

- **A fixed distance of $3$ cm from a fixed point $O$** — a circle with centre $O$ and radius $3$ cm
- **A fixed distance of $2$ cm from a straight line** — a pair of lines parallel to it, $2$ cm away on either side
- Equidistant from two parallel lines — the parallel line midway between them

Worked example 1. A bicycle wheel of radius $35$ cm rolls along a flat road. The locus of its centre is a straight line parallel to the road, $35$ cm above it.

Worked example 2. A goat is tied to a peg with a $5$ m rope. The boundary of the grass it can reach is a circle of radius $5$ m, enclosing an area of

$$\pi r^2=3.14\times 5^2=78.5{\text{ m}}^2$$

An everyday example. A tethered goat grazes a neat circular patch.

The substance. A locus must include every point that fits and exclude every point that does not, so a single line on one side of a road is not the full answer.

The locus of points at a fixed distance from a fixed point is a circle, the locus of points equidistant from two fixed points is the perpendicular bisector of the segment joining them, and the locus of points equidistant from two intersecting lines is the pair of lines bisecting the angles between them.

1. Fixed distance from a fixed point — a circle.

**2. Equidistant from two points $A$ and $B$** — the perpendicular bisector of $AB$.

Why. If $P$ lies on the perpendicular bisector through the mid-point $M$, then triangles $PMA$ and $PMB$ share $PM$, have $AM=BM$ and right angles at $M$, so they are congruent by SAS and $PA=PB$.

3. Equidistant from two intersecting lines — the two angle bisectors, which are perpendicular to each other.

Worked example. $AB=8$ cm. A point $P$ on the perpendicular bisector of $AB$ is $3$ cm from the mid-point $M$.

$$PA=PB=\sqrt{4^2+3^2}=\sqrt{25}=5\text{ cm}$$

An everyday example. A water tank meant to be equally far from two villages must stand somewhere on the perpendicular bisector of the line joining them.

The substance. Two intersecting lines have two angle bisectors, so the third locus is a pair of lines, not just one.

For a perpendicular bisector, draw equal arcs from both ends of the segment and join their crossing points; for an angle bisector, mark equal distances on both arms, draw equal arcs from those marks and join the vertex to where the arcs cross.

**Perpendicular bisector of $AB$:

- Open the compasses to more than half of $AB$
- With centre $A$**, draw arcs above and below $AB$
- **With centre $B$ and the same radius**, draw arcs cutting the first two at $P$ and $Q$
- **Join $PQ$** — it bisects $AB$ at right angles

**Bisector of $\angle ABC$:

- With centre $B$**, draw an arc cutting $BA$ at $P$ and $BC$ at $Q$
- **With centres $P$ and $Q$ and equal radii**, draw arcs crossing at $R$
- **Join $BR$ — it bisects the angle

Worked example.** Constructing $60^{\circ }$ and bisecting it gives $30^{\circ }$; bisecting a $70^{\circ }$ angle gives two angles of $35^{\circ }$ each, which can be checked with a protractor.

An everyday example. A carpenter marks the centre line of a plank with the same equal arcs.

The substance. **The radius must be more than half of $AB$**; otherwise the arcs from $A$ and $B$ never meet.

Construct each locus separately and accurately; the points where the two loci meet are the points satisfying both conditions, and you then measure the distance asked for.

Worked example 1. $AB=6$ cm. Find the points that are $4$ cm from $A$ and equidistant from $A$ and $B$.

- Locus 1: circle with centre $A$, radius $4$ cm
- Locus 2: perpendicular bisector of $AB$
- They meet at two points $P$ and $Q$, one on each side of $AB$

By calculation, each point is $\sqrt{4^2-3^2}=\sqrt{7}\approx 2.65$ cm from the mid-point of $AB$, so

$$PQ=2\sqrt{7}\approx 5.29\text{ cm}$$

which your measured answer should match closely.

Worked example 2. In $△ABC$, find the point equidistant from $AB$ and $BC$ and also equidistant from $B$ and $C$: construct the **bisector of $\angle B$ and the perpendicular bisector of $BC$, and mark where they meet.

An everyday example. Planners siting a school equally far from two villages and near a main road use this idea.

The boundary case. With a $2$ cm circle instead of $4$ cm, there is no such point**, because the perpendicular bisector is $3$ cm from $A$.