Loci

Say the words "stay the same distance from that tree" out loud, and you have already described a locus — the invisible path traced by every point obeying a rule. Loci convert a spoken condition into an exact geometric shape: sometimes a circle, sometimes a straight line, always something you can draw with nothing but a ruler and compass. Class X restricts you to three named loci — fixed distance from a point, equidistant from two points, and equidistant from two lines — and asks you to state them, construct them accurately, and combine two of them at once to pin down a single point or a small set of points. No formal proofs are demanded, but knowing why each locus takes the shape it does makes the constructions far harder to forget under exam pressure. This article keeps the same three loci and the same worked numbers as your syllabus notes, then adds the reasoning behind the perpendicular-bisector locus, a fully computed compound-locus example, and a quick exam-card summary.

The core insight

A locus is the complete set of points satisfying one stated geometric condition, and compound-locus problems are solved by constructing two such loci separately and marking where they intersect.

The core idea

A locus is the set of all points — no more, no fewer — that satisfy a given rule. Two things must always accompany a locus statement: the shape, and the condition that produced it. Saying "it's a circle" without saying centre and radius is an incomplete answer.

1. Fixed distance from a fixed point. For a fixed point $O$ and constant $r$ , every point $P$ with $OP = r$ traces a circle.

\{P : OP = r\} = \text{circle, centre } O, \text{ radius } r

where: $O$ is the fixed centre, $r$ is the fixed radius, and $P$ ranges over every point in the plane satisfying the distance condition.

2. Equidistant from two given points. For fixed points $A$ and $B$ , every point $P$ with $PA = PB$ lies on one specific straight line.

\{P : PA = PB\} = \text{perpendicular bisector of } AB

where: $AB$ is the segment joining the two fixed points, and the perpendicular bisector is the line through the midpoint of $AB$ at right angles to it.

3. Equidistant from two intersecting lines. Every point equidistant from two lines crossing at $O$ lies on one of the two angle bisectors of the angles formed at $O$ — and because vertically opposite angles come in pairs, there are always two such bisector rays, and the two bisectors are perpendicular to each other. These three are not independent party tricks — they are the only tools you need. Every compound-locus question in the syllabus reduces to drawing two of these three shapes and marking the point(s) common to both.

Figure

Three panels showing a circle locus around a point, a straight perpendicular bisector locus between two points, and a pair of perpendicular angle bisector rays at the crossing of two lines

Derivation AWhy the Perpendicular Bisector Locus Works

Show, using only the fact that two right-angled triangles sharing a common leg and a pair of equal hypotenuses must be congruent, that a point $P$ satisfies $PA = PB$ for two fixed points $A$ and $B$ if and only if $P$ lies exactly on the perpendicular line through the midpoint $M$ of segment $AB$ , which is precisely the claim used without proof in every compound-locus construction on this locus.

Step 1 — Set up the midpoint and the candidate line. Let $M$ be the midpoint of $AB$ , so $AM = MB$ . Let $\ell$ be the line through $M$ perpendicular to $AB$ . We must show two things: every point on $\ell$ is equidistant from $A$ and $B$ , and every point equidistant from $A$ and $B$ lies on $\ell$ .

Step 2 — Points on $\ell$ are equidistant (the easy direction). Take any point $P$ on $\ell$ , $P \ne M$ . Triangles $PMA$ and $PMB$ share the leg $PM$ , have $AM = MB$ by construction, and both angles at $M$ are $90^\circ$ since $\ell \perp AB$ . By the SAS congruence rule these two triangles are congruent, so the third sides are equal: $PA = PB$ . If $P = M$ itself, $MA = MB$ trivially. So every point of $\ell$ satisfies the condition.

Step 3 — Equidistant points lie on $\ell$ (the converse). Now suppose $PA = PB$ for some point $P$ not assumed to be on $\ell$ . Drop the foot of the perpendicular from $P$ to line $AB$ and call it $F$ . Using the Pythagorean theorem in right triangles $PFA$ and $PFB$ :

PA^2 = PF^2 + FA^2, \qquad PB^2 = PF^2 + FB^2

where: $PF$ is the common perpendicular distance from $P$ to line $AB$ , and $FA$ , $FB$ are the distances from $F$ to $A$ and $B$ along that line. Since $PA = PB$ , the two right-hand sides are equal, so $FA^2 = FB^2$ , giving $FA = FB$ — meaning $F$ is equidistant from $A$ and $B$ along the line $AB$ , which forces $F = M$ . So $P$ lies directly above (or below) $M$ on the perpendicular through $M$ — that is, $P$ lies on $\ell$ .

Step 4 — Conclude. Both directions hold, so $\{P : PA = PB\}$ is exactly the line $\ell$ , no more and no less — confirming the locus stated in the core idea, and the same congruent-triangle idea (equal legs, shared side, right angle) is what makes the angle-bisector locus and compass constructions trustworthy too.

Worked example

Circle of radius $5\ \text{cm}$ centred at $O$ is drawn, and $AB = 6\ \text{cm}$ is a segment whose perpendicular bisector passes through $O$ , with $O$ lying $3\ \text{cm}$ from the midpoint $M$ of $AB$ . Find the distance from each point $P$ satisfying both $OP = 5\ \text{cm}$ and $PA = PB$ to the points $A$ and $B$ .

Step 1 — Identify the two loci. Locus 1 is the circle $OP = 5\ \text{cm}$ (the fixed-distance locus). Locus 2 is the perpendicular bisector of $AB$ , the line $PA = PB$ . Since $O$ is given to lie on this perpendicular bisector, the entire line 2 passes straight through the centre of circle 1 — so the circle and the line meet at exactly two points, one on each side of $O$ along the line.

Step 2 — Locate the two intersection points along the bisector line. Along the perpendicular bisector, $M$ sits $3\ \text{cm}$ from $O$ . A point $P$ on the circle is $5\ \text{cm}$ from $O$ measured along this same line, so there are two cases: $P_1$ on the far side of $M$ from $O$ , at distance $PM_1 = 5 - 3 = 2\ \text{cm}$ from $M$ ; and $P_2$ on the same side as $O$ but beyond it, at distance $PM_2 = 5 + 3 = 8\ \text{cm}$ from $M$ .

Step 3 — Apply the Pythagorean relation from the derivation. Because $M$ is the midpoint of $AB = 6\ \text{cm}$ , $MA = MB = 3\ \text{cm}$ . For any point $P$ on the perpendicular bisector, $PA = PB = \sqrt{PM^2 + MA^2}$ — exactly the right-triangle relation used in Step 3 of the derivation above.

Step 4 — Compute both answers. For $P_1$ : $PA = PB = \sqrt{2^2 + 3^2} = \sqrt{13} \approx 3.61\ \text{cm}$ . For $P_2$ : $PA = PB = \sqrt{8^2 + 3^2} = \sqrt{73} \approx 8.54\ \text{cm}$ .

The compound locus has exactly two solution points, with $PA = PB \approx 3.61\ \text{cm}$ for the nearer one and $PA = PB \approx 8.54\ \text{cm}$ for the farther one.

Competitive stretch

If instead $O$ were not on the perpendicular bisector of $AB$ (a more general compound-locus setup), the circle and the bisector line would generally still meet in at most two points, but you could no longer measure distances along a single shared line — you would need to set up coordinates, write the circle's equation and the bisector's equation, and solve them simultaneously. The construction method (draw both loci, mark the intersection) stays identical either way; only the arithmetic to find exact distances changes.

Common traps

Naming the shape of a locus without stating the defining condition (centre and radius, or the two points/lines involved) — a locus answer is incomplete without both.
Forgetting that two intersecting lines give two perpendicular angle bisectors, not one — "equidistant from two lines" means the pair, unless the question restricts to a single angle.
In compound-locus problems, constructing only one locus and stopping, instead of finding where both loci intersect.
Mixing up "equidistant from two points" (perpendicular bisector of the segment) with "equidistant from two lines" (angle bisector) — these use different constructions for different conditions.

Ask the tutor: Stuck picturing how a circle and a perpendicular bisector intersect in a compound-locus problem? Ask the AI tutor to walk through the construction with you step by step.

Do I need to prove why the perpendicular bisector is the correct locus in the exam?

No — the syllabus only asks you to state each of the three standard loci and construct them correctly with ruler and compass; formal proofs are explicitly not required. The congruent-triangle reasoning shown above is there so the rule stops feeling arbitrary and becomes easy to recall, not because you'll be asked to reproduce it as a proof.

In a compound locus question, how many points can satisfy both conditions?

It depends entirely on how the two loci are arranged. A circle and a line can meet at two points, one point (if the line is tangent to the circle), or zero points (if the line misses the circle entirely) — always check your construction rather than assuming there are always two solutions, as in the worked example above.

Why is the locus of "equidistant from two points" a full line and not just a segment?

Because the defining condition PA = PB only involves distances, and it turns out (as shown in the derivation) that every point directly above or below the midpoint M on the perpendicular line satisfies this equally, no matter how far from M it is. There is nothing in the condition that limits how far along that perpendicular direction P can go, so the locus extends infinitely in both directions as a full line, not a bounded segment.

Loci

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads