Chapter 8: Constructions and Loci
Geometric constructions use only a compass and a straight-edged ruler — no measurements. Accurate constructions are foundational skills that also underpin the theory of loci. A locus is the set of all points satisfying a given geometric condition; it can be a line, a circle, or a region.
Exam Board Coverage
All constructions and loci topics appear at both Foundation and Extended tier. Compound loci (intersecting two or more loci to find a region or a set of points) are more common at Extended. Always leave your construction arcs visible — rubbing them out causes loss of marks even if the final line is correct.
8.1 Equipment and Conventions
Tools for Construction
- Compass: used to draw arcs and circles of a fixed radius. Always keep the compass at a constant opening when drawing arcs for a single construction step.
- Ruler (straight edge): used only to draw straight lines — you may not use the ruler's markings to measure lengths during a construction unless the question asks you to.
- Protractor: used to measure and draw angles (not part of a pure construction but used in "draw accurately" questions).
Golden Rule
Never erase construction arcs. Examiners look for arcs as evidence that you used a compass. If arcs are missing, method marks may not be awarded even if your final answer is correct.
8.2 Perpendicular Bisector of a Line Segment
The perpendicular bisector of $AB$ is the line that is perpendicular to $AB$ and passes through its midpoint. Every point on the perpendicular bisector is equidistant from $A$ and $B$.
Method: Perpendicular Bisector
- Open the compass to a radius greater than half the length of $AB$ (more than $\tfrac{1}{2}AB$).
- With the compass point on $A$, draw an arc above and below the line.
- Without changing the compass radius, place the point on $B$ and draw arcs that intersect the first pair of arcs at two points $P$ and $Q$.
- Draw the straight line through $P$ and $Q$ — this is the perpendicular bisector.
Example 8.1 — Perpendicular bisector and equidistance
Points $A$ and $B$ are $8\,\text{cm}$ apart. Describe the locus of all points equidistant from $A$ and $B$.
The locus is the perpendicular bisector of $AB$. Construct it using the method above. Every point on this line is exactly $\tfrac{1}{2}AB = 4\,\text{cm}$ from both $A$ and $B$.
8.3 Angle Bisector
The angle bisector of $\angle AOB$ is the ray from $O$ that cuts the angle exactly in half. Every point on the angle bisector is equidistant from the two lines forming the angle.
Method: Angle Bisector
- Place the compass point at the vertex $O$ and draw an arc that crosses both arms of the angle at points $P$ and $Q$.
- Without changing the compass radius, place the point on $P$ and draw an arc in the interior of the angle.
- With the same radius, place the point on $Q$ and draw an arc that intersects the previous one at $R$.
- Draw the ray from $O$ through $R$ — this is the angle bisector.
Example 8.2 — Angle bisector
Two roads meet at a junction forming an angle of $70°$. A new lamppost is to be placed equidistant from both roads. Where should it be placed?
The lamppost must lie on the angle bisector of the $70°$ angle. Construct the angle bisector: the set of all equidistant points from both roads.
8.4 Perpendicular from a Point to a Line
Method: Perpendicular from Point $P$ to Line $\ell$
- Place the compass at $P$ and draw an arc that crosses $\ell$ at two points $X$ and $Y$ (choose a large enough radius).
- Now construct the perpendicular bisector of $XY$ — this passes through $P$ and is perpendicular to $\ell$.
- The foot of the perpendicular (where this line meets $\ell$) is the closest point on $\ell$ to $P$.
8.5 Constructing Standard Angles
Standard Angle Constructions
- $60°$ angle: Construct an equilateral triangle. Place the compass at one endpoint, draw an arc with radius equal to the line length, then mark the intersection — the angle at each vertex is $60°$.
- $90°$ angle: Construct the perpendicular bisector of any segment, or use the method for a perpendicular from a point on a line. The perpendicular meets the line at $90°$.
- $45°$ angle: Construct a $90°$ angle and then bisect it using the angle bisector method.
- $30°$ angle: Construct a $60°$ angle and bisect it.
8.6 Constructing Triangles
Constructing a Triangle from SSS (all three sides known)
- Draw the base $AB$ of the correct length using a ruler.
- Set the compass radius to $AC$ and draw an arc centred at $A$.
- Set the compass radius to $BC$ and draw an arc centred at $B$.
- The intersection of the two arcs is vertex $C$. Join $AC$ and $BC$.
Constructing a Triangle from SAS (two sides and included angle)
- Draw side $AB$ of the correct length.
- Construct angle $A$ using a protractor.
- Mark off the length $AC$ along the angle ray.
- Join $B$ to $C$.
Constructing a Triangle from ASA (two angles and included side)
- Draw side $AB$ of the correct length.
- Construct angle $A$ at vertex $A$, and angle $B$ at vertex $B$.
- Extend the two angle rays until they intersect — that intersection is $C$.
Example 8.3 — Construct triangle SSS
Construct triangle $ABC$ with $AB = 7\,\text{cm}$, $BC = 5\,\text{cm}$, $CA = 6\,\text{cm}$.
Step 1 Draw $AB = 7\,\text{cm}$.
Step 2 Set compass to $6\,\text{cm}$, centre $A$; draw arc above $AB$.
Step 3 Set compass to $5\,\text{cm}$, centre $B$; draw arc intersecting the first arc at $C$.
Step 4 Join $AC$ and $BC$. Leave all arcs visible.
8.7 Introduction to Loci
Locus (plural: Loci)
A locus is the set of all points that satisfy a given geometric condition. Think of it as the path traced by a moving point that obeys a rule at all times.
For example: "the locus of points $2\,\text{cm}$ from point $A$" is a circle of radius $2\,\text{cm}$ centred at $A$.
In exam questions, a locus is usually described verbally and you must:
- Identify the correct geometric shape (circle, line, parallel lines, angle bisector, …).
- Draw or describe it accurately on the diagram.
- For regions: shade or clearly label the required area.
8.8 The Four Standard Loci
| Condition | Locus | Notes |
|---|---|---|
| Fixed distance $r$ from a point $P$ | Circle, centre $P$, radius $r$ | The set of all points exactly $r$ from $P$ |
| Equidistant from two fixed points $A$ and $B$ | Perpendicular bisector of $AB$ | Construct using the perpendicular bisector method |
| Fixed distance $d$ from a straight line $\ell$ | Two lines parallel to $\ell$, each at distance $d$ (one on each side) | For a line segment, add semicircles at each end |
| Equidistant from two intersecting lines $\ell_1$ and $\ell_2$ | The two angle bisectors of the angles formed by $\ell_1$ and $\ell_2$ | There are always two perpendicular angle bisectors |
Example 8.4 — Describing a locus
Point $P$ is inside an equilateral triangle of side $6\,\text{cm}$. Describe the locus of all points inside the triangle that are closer to side $AB$ than to either $AC$ or $BC$.
Answer The region is bounded by the two angle bisectors at vertices $A$ and $B$ (equidistant from the two pairs of sides). The locus is the interior region of the triangle that lies below both bisectors — it is a triangular sub-region with vertex at the intersection of the two bisectors (the incentre) and base along $AB$.
8.9 Compound Loci Extended
A compound locus involves two or more conditions simultaneously. The solution is the intersection of the individual loci.
Method for Compound Loci
- Read each condition separately.
- Draw each locus on the same diagram.
- Find the intersection points (or region) that satisfies all conditions at once.
- Mark or shade clearly and label.
Example 8.5 — Two-condition locus
Points $A$ and $B$ are $8\,\text{cm}$ apart. Find all points that are: (i) $5\,\text{cm}$ from $A$, and (ii) $5\,\text{cm}$ from $B$.
Step 1 Locus for (i): circle centred at $A$, radius $5\,\text{cm}$.
Step 2 Locus for (ii): circle centred at $B$, radius $5\,\text{cm}$.
Step 3 The circles intersect at two points $P$ and $Q$ (above and below $AB$). These two points satisfy both conditions.
Note: $AB = 8 < 5+5 = 10$, so the circles do intersect. We can verify: $P$ lies on the perpendicular bisector of $AB$ at a distance of $\sqrt{5^2 - 4^2} = \sqrt{9} = 3\,\text{cm}$ from $AB$.
Example 8.6 — Region satisfying two inequalities
In a rectangle $ABCD$ with $AB = 10\,\text{cm}$ and $BC = 6\,\text{cm}$, find and shade the region inside the rectangle that satisfies both:
- Within $4\,\text{cm}$ of $A$
- Closer to $AD$ than to $BC$
Step 1 Draw a quarter-circle of radius $4\,\text{cm}$ centred at $A$ (inside the rectangle).
Step 2 Draw the perpendicular bisector of $AB$ — a vertical line at $x = 5\,\text{cm}$ from $AD$. Points closer to $AD$ are on the $AD$ side of this line.
Step 3 Shade the region that is inside the quarter-circle AND to the left of the perpendicular bisector. This is a quarter-circle arc region.
Exercises
Exercise 8.1
Describe, in words, the locus of all points that are exactly $3\,\text{cm}$ from a fixed point $O$.
Show Solution
A circle with centre $O$ and radius $3\,\text{cm}$.
Exercise 8.2
Two parallel lines are $6\,\text{cm}$ apart. Describe the locus of all points equidistant from both lines.
Show Solution
A straight line parallel to both, halfway between them — at distance $3\,\text{cm}$ from each line.
Exercise 8.3
Describe the locus of all points $4\,\text{cm}$ from a line segment $PQ$ of length $10\,\text{cm}$. (Include points beyond the endpoints.)
Show Solution
Two parallel lines $4\,\text{cm}$ from $PQ$ (one on each side), joined at each end by a semicircle of radius $4\,\text{cm}$ centred at $P$ and $Q$ respectively. The shape is a "stadium" or "discorectangle".
Exercise 8.4
Explain why the perpendicular bisector of $AB$ is the locus of all points equidistant from $A$ and $B$.
Show Solution
Let $M$ be the midpoint of $AB$ and $P$ be any point on the perpendicular bisector. Triangle $APM \cong$ triangle $BPM$ (SAS: $AM = BM$, $\angle PMA = \angle PMB = 90°$, $PM$ common). Therefore $PA = PB$. Conversely, if $PA = PB$, the locus by symmetry lies on the perpendicular bisector.
Exercise 8.5
State the two conditions needed for a point $P$ to lie on the angle bisector of $\angle AOB$.
Show Solution
$P$ lies on the angle bisector of $\angle AOB$ if and only if $P$ is (1) in the interior of the angle and (2) equidistant from both arms $OA$ and $OB$.
Exercise 8.6
Construct triangle $ABC$ with $AB = 6\,\text{cm}$, $BC = 5\,\text{cm}$, $AC = 7\,\text{cm}$. (Describe the steps.)
Show Solution
1. Draw $AB = 6\,\text{cm}$. 2. Arc, radius $7\,\text{cm}$, centre $A$. 3. Arc, radius $5\,\text{cm}$, centre $B$. 4. Mark intersection as $C$. 5. Join $AC$ and $BC$. Leave arcs visible.
Exercise 8.7 Extended
Points $A$ and $B$ are $6\,\text{cm}$ apart. Find all points that are simultaneously $4\,\text{cm}$ from $A$ and $4\,\text{cm}$ from $B$. How many such points are there? Explain.
Show Solution
Draw circle $A$ (radius 4) and circle $B$ (radius 4). Since $AB = 6 < 4+4 = 8$ and $AB = 6 > |4-4| = 0$, the circles intersect at two points — one above and one below $AB$. Each intersection point is at distance $\sqrt{4^2-3^2} = \sqrt{7}\,\text{cm}$ from line $AB$.
Exercise 8.8 Extended
In triangle $PQR$, a point $X$ must satisfy: (1) $X$ is closer to $PQ$ than to $QR$, and (2) $X$ is closer to $P$ than to $R$. Describe how to find and shade the required region.
Show Solution
Condition (1): Construct the angle bisector of $\angle PQR$ — this separates the triangle into the region closer to $PQ$ (upper part) and closer to $QR$ (lower part).
Condition (2): Construct the perpendicular bisector of $PR$ — this separates the triangle into the region closer to $P$ (left part) and closer to $R$ (right part).
The required region is the intersection of both parts — shade the area that is above the angle bisector and to the left of the perpendicular bisector of $PR$.
Exam Tips
Tip 1 — Never erase arcs
Construction arcs are your evidence. An examiner cannot award method marks without seeing them, even if your final line is in exactly the right place. Leave all arcs visible throughout.
Tip 2 — State the locus type before drawing
Read the locus condition, identify which of the four standard types it is, then draw. Trying to "guess" the shape from the diagram is slow and error-prone.
Tip 3 — Locus of fixed distance from a segment includes the ends
If asked for the locus of points within $r$ of a line segment (not an infinite line), remember to include the semicircular end caps. A common mistake is drawing only the two parallel lines and forgetting the rounded ends.
Tip 4 — Shade the correct region explicitly
For compound loci, identify each boundary line, then decide which side of each boundary you need. Use light shading and label the region or mark a representative point inside it. When the question says "closer to X than to Y", the boundary is the perpendicular bisector (for points) or the angle bisector (for lines).