Loci and Constructions
A locus (plural: loci) is the set of all points satisfying a given geometric condition. Constructions use only a ruler and compasses. Mastering both is essential for map-work, proof, and multi-constraint problems at Grade 9.
Learning Objectives
- Construct perpendicular bisectors and angle bisectors accurately using compasses
- Describe loci for the four standard conditions and combine them algebraically
- Find regions satisfying multiple simultaneous locus conditions
- Apply loci to scale drawings, maps, and bearing problems
- Construct the perpendicular from a point to a line and use constructions in proof
🔑 Core Concepts
2.1 What is a Locus?
A locus is a path or region traced by every point that satisfies a particular rule. Think of it as "all points for which something is true." The shape of a locus depends entirely on the constraint given.
2.2 Locus 1 — Fixed Distance from a Fixed Point
Condition: All points exactly $r$ units from a fixed point $P$.
Why it's a circle: By the definition of a circle — every point on a circle is the same distance (the radius) from the centre. So this locus is the definition of a circle.
Construction method: Set compasses to radius $r$. Place point on $P$. Draw a complete circle.
2.3 Locus 2 — Equidistant from Two Fixed Points
Condition: All points equidistant from two fixed points $A$ and $B$.
Why it's the perpendicular bisector: Any point equidistant from $A$ and $B$ lies on the unique line that cuts $AB$ at right angles through its midpoint. This is proven by congruent triangles (SSS or SAS).
Construction method (perpendicular bisector):
2.4 Locus 3 — Equidistant from Two Intersecting Lines
Condition: All points equidistant from two lines (or line segments) that intersect at a point $O$.
Why it's the angle bisector: The perpendicular distance from a point to a line is minimised along the perpendicular. The set of points at equal perpendicular distances from both lines forms the angle bisector. There are actually two angle bisectors (for the two pairs of vertical angles), but questions usually ask for the one inside a given angle.
Construction method (angle bisector):
2.5 Locus 4 — Fixed Distance from a Line Segment
Condition: All points exactly $d$ units from a line segment $AB$.
Why it's a "stadium" shape: Along the length of $AB$, points at distance $d$ are parallel lines offset by $d$ on each side. At the endpoints $A$ and $B$, the nearest point on the segment is the endpoint itself, so those points form semicircles of radius $d$ centred on $A$ and $B$.
2.6 Perpendicular from an External Point to a Line
To drop a perpendicular from point $P$ to line $\ell$:
2.7 Intersection of Loci — Compound Conditions
Grade 9 problems typically require the intersection of two or more loci. The solution region is the set of points satisfying all conditions simultaneously. You must construct each locus separately, then identify the overlap.
2.8 Scale Drawings and Bearings
In scale drawing problems, loci represent real-world boundaries. A scale of $1:n$ means $1$ cm on paper represents $n$ cm in reality.
🗺️ Visual Notes
Constructions
- Fixed distance from point → circle
- Equidistant from 2 points → perp. bisector
- Equidistant from 2 lines → angle bisector
- Fixed distance from segment → stadium shape
- Perpendicular bisector (compasses, no ruler measure)
- Angle bisector (equal arcs from vertex)
- Perpendicular from point to line
- Circumcircle (3 perp. bisectors meet)
- Intersection = points satisfying all conditions
- Inequalities define half-plane regions
- Shade valid regions carefully
- State which boundary is included/excluded
- Convert real distances using scale factor
- Bearings: clockwise from North, 3 digits
- Draw loci to scale, shade valid region
- Convert answer back to real units
- Circle: $(x-a)^2+(y-b)^2=r^2$
- Closer to $A$: $PA < PB$
- Within distance: $PA \le r$
- Locus = satisfies equality; region = inequality
- Leave all arcs visible
- Label every locus you draw
- Shade regions with light pencil
- Check construction with ruler at end
Comparison of the Four Standard Loci
| Condition | Resulting Locus | Construction Tool | Key Property |
|---|---|---|---|
| Exactly $r$ from fixed point $P$ | Circle, centre $P$, radius $r$ | Compasses set to $r$ | All points equidistant from $P$ |
| Equidistant from two points $A$, $B$ | Perpendicular bisector of $AB$ | Compasses + ruler | Meets $AB$ at 90° at its midpoint |
| Equidistant from two lines $\ell_1$, $\ell_2$ | Angle bisector | Compasses only | Bisects the angle between the lines |
| Exactly $d$ from line segment $AB$ | Two parallel lines + two semicircles | Ruler + compasses | "Stadium" shape; semicircles at endpoints |
Decision Tree: Which Locus to Construct?
→ Circle
→ Perp. Bisector
→ Angle Bisector
→ Stadium Shape
Regions vs Boundaries: Algebraic Descriptions
| English Condition | Algebraic Form | Drawn as | Boundary Included? |
|---|---|---|---|
| Exactly $r$ from $P$ | $(x-a)^2+(y-b)^2 = r^2$ | Circle (line only) | Yes — draw solid line |
| Within $r$ of $P$ | $(x-a)^2+(y-b)^2 \leq r^2$ | Disc (shaded interior) | Yes — solid boundary |
| Closer to $A$ than to $B$ | $PA < PB$ | Half-plane (shade $A$'s side) | No — dashed perp. bisector |
| Closer to line $\ell_1$ than $\ell_2$ | $d(P,\ell_1) < d(P,\ell_2)$ | Half-angle region | No — dashed angle bisector |
| At least $d$ from segment $AB$ | $d(P, AB) \geq d$ | Exterior of stadium | Yes — solid stadium boundary |
✏️ Worked Examples
- closer to $A$ than to $B$, and
- within 5 cm of $B$.
Condition 2: "Within 5 cm of $B$" → The boundary is a circle of radius 5 cm centred at $B$. The interior satisfies $PB \leq 5$.
- within 8 km of the coastguard station $C$,
- more than 5 km from the lighthouse $L$,
- closer to $L$ than to $C$.
Locus A: "Within 8 km of $C$" → circle centre $C$, radius $8 \div 2 = 4$ cm.
Locus B: "More than 5 km from $L$" → exterior of circle centre $L$, radius $5 \div 2 = 2.5$ cm.
Locus C: "Closer to $L$ than to $C$" → $L$-side of perpendicular bisector of $CL$.
• Inside the 4 cm circle (Locus A)
• Outside the 2.5 cm circle (Locus B)
• On the $L$-side of the perpendicular bisector (Locus C)
Shade this region (a "crescent-like" area to the right of the perp. bisector, inside the large circle, outside the small circle).
❓ Exam Questions
State the locus of all points that are exactly 4 cm from a fixed point $P$.
Mark scheme: 1 mark for "circle, centre $P$, radius 4 cm" (all three components required).
Construct the perpendicular bisector of a line segment $AB$ where $AB = 7$ cm. Mark clearly the midpoint $M$ of $AB$.
B1 — Correct arcs: compasses set to more than half $AB$, arcs drawn from both $A$ and $B$, crossing above and below the line (arcs must be visible).
B1 — Straight line drawn through the two arc intersections, perpendicular to $AB$; midpoint $M$ marked where bisector crosses $AB$.
A goat is tethered by a 6 m rope to a corner of a 4 m × 3 m rectangular barn. The goat can graze anywhere the rope reaches, but cannot pass through the barn. Describe and sketch the locus of the maximum extent of grazing area. [Hint: consider what happens when the rope wraps around the corner.]
The goat starts at corner $C$ with 6 m rope. It can graze a three-quarter circle of radius 6 m on the open side (270° arc, since the barn occupies 90° of the space).
When the rope wraps around the adjacent corner $D$ (3 m away), there is $6 - 3 = 3$ m remaining. This allows a quarter-circle of radius 3 m around corner $D$ (90° arc, into the space around that side of the barn).
Similarly, if the rope wraps around the other adjacent corner $E$ (4 m away), $6 - 4 = 2$ m remains. This allows a quarter-circle of radius 2 m around corner $E$.
Mark scheme:
M1 — Three-quarter circle of radius 6 m identified at tethering corner.
M1 — At least one wrapped corner with correct remaining rope length (3 m or 2 m).
A1 — Both quarter circles at adjacent corners correctly drawn/described (radii 3 m and 2 m).
Two towns, $A$ and $B$, are 10 km apart. A mobile phone mast can be placed anywhere satisfying all of the following:
(i) within 7 km of $A$
(ii) within 6 km of $B$
(iii) closer to $B$ than to $A$
Using a scale of 1 cm : 2 km, construct the region where the mast can be placed. State the types of boundaries used.
Boundary 1: Circle centre $A$, radius 3.5 cm (solid — includes 7 km boundary). Mast inside. ✓ B1
Boundary 2: Circle centre $B$, radius 3 cm (solid — includes 6 km boundary). Mast inside. ✓ B1
Boundary 3: Perpendicular bisector of $AB$ at 2.5 cm from each end, drawn as dashed line. Mast on $B$-side (strictly closer to $B$). ✓ B1
Shading: Region where all three hold — intersection of both circle interiors on the $B$-side of the perpendicular bisector. ✓ B1
Accept solid or dashed for the perpendicular bisector if student correctly states whether the boundary is included.
Point $P$ has coordinates $(2, 3)$ and point $Q$ has coordinates $(8, 3)$. Write down the equation of the locus of all points equidistant from $P$ and $Q$. Show your working.
The locus equidistant from $P(2,3)$ and $Q(8,3)$ is the perpendicular bisector of $PQ$.
Midpoint of $PQ$: $M = \left(\frac{2+8}{2}, \frac{3+3}{2}\right) = (5, 3)$. (M1)
Gradient of $PQ = \frac{3-3}{8-2} = 0$ (horizontal line).
Perpendicular bisector is vertical through $(5, 3)$. (M1)
Equation: $x = 5$ (A1)
Alternative method: Set $\sqrt{(x-2)^2+(y-3)^2} = \sqrt{(x-8)^2+(y-3)^2}$, square and expand to get $x=5$. Full marks if correct.
Grade 9 Challenge: A treasure is buried at a point $T$ that satisfies all of the following conditions on a map (scale 1 cm : 50 m):
(i) $T$ is equidistant from landmarks $X$ and $Y$ (which are 8 cm apart on the map).
(ii) $T$ is within 5 cm of $X$ on the map.
(iii) $T$ is closer to line $\ell_1$ than to line $\ell_2$, where $\ell_1$ and $\ell_2$ meet at a point on the perpendicular bisector of $XY$ at angle 60°.
Construct the valid region for $T$ and find the real-world distance from $X$ to the nearest point of this region.
B1 — Perpendicular bisector of $XY$ correctly constructed (compasses set > 4 cm, arcs from both $X$ and $Y$, line through intersections).
B1 — Circle centre $X$, radius 5 cm drawn (solid boundary).
B1 — Angle bisector of the 60° angle at the given point on the perpendicular bisector (dashed line, since condition is strict).
B1 — Valid region shaded: on the perpendicular bisector (or the line itself is the boundary for condition i), inside the circle, on $\ell_1$'s side of the angle bisector.
M1 — Nearest point to $X$ on the valid region identified: this is where the perpendicular bisector meets the boundary of the circle centred at $X$, at distance 5 cm.
A1 — Real distance: $5 \text{ cm} \times 50 = 250 \text{ m}$. Answer: 250 m.
Note: Condition (i) means $T$ is ON the perpendicular bisector, so it is a line, not a region — full marks if student correctly identifies this is a 1D locus intersected with the circle and half-angle region.
⭐ Grade 9 Model Answers
Full annotated answer to Question 6 — the Grade 9 multi-constraint scale drawing problem.
Step 1 — Identify what each condition produces
Condition (i): "$T$ equidistant from $X$ and $Y$" → This is not a region but a specific line: the perpendicular bisector of $XY$. $T$ must lie on this line exactly. This immediately reduces the search from a 2D region to a 1D line.
Why this earns marks: Many students treat this as a region ("near the midpoint"). A Grade 9 student recognises that equidistance from two points produces a unique line, not a band.
Step 2 — Intersect with Condition (ii)
Condition (ii): "$T$ within 5 cm of $X$" → Circle, centre $X$, radius 5 cm. Combined with condition (i), $T$ must be on the perpendicular bisector of $XY$ AND inside this circle. This restricts $T$ to the segment of the perpendicular bisector inside the circle.
Constructing this segment: The perpendicular bisector meets the circle at two points. The valid portion is the chord of the perpendicular bisector inside the circle.
Step 3 — Apply Condition (iii)
Condition (iii): "$T$ closer to $\ell_1$ than $\ell_2$" → The boundary is the angle bisector of the 60° angle. Points on $\ell_1$'s side satisfy $d(T,\ell_1) < d(T,\ell_2)$. This halves the remaining chord segment: $T$ is on the correct side of the angle bisector.
Step 4 — Nearest point to X
The nearest point of the valid region to $X$ is where the perpendicular bisector first enters the region. Since the perpendicular bisector is at equal distance from $X$ and $Y$ (i.e., at distance 4 cm from each, as $XY = 8$ cm), the nearest point on the perpendicular bisector to $X$ is the midpoint $M$, at 4 cm from $X$.
But the valid region could be further restricted by condition (iii). If the angle bisector cuts the perpendicular bisector segment, then the nearest point in the region is the intersection of the angle bisector with the perpendicular bisector segment.
In this problem, assuming the nearest boundary point is the midpoint $M$ (at 4 cm from $X$): Real distance = $4 \times 50 = 200$ m.
If the constraint from condition (iii) means the region starts at 5 cm from $X$ (the circle boundary): Real distance = $5 \times 50 = 250$ m.
Mark Allocation Summary:
- B1: Perpendicular bisector correctly constructed with visible arcs
- B1: Circle, centre $X$, radius 5 cm, solid boundary
- B1: Angle bisector constructed at the specified point, dashed (strict inequality)
- B1: Correct region identified and shaded (intersection of all three)
- M1: Nearest point method shown — identifies the point on the boundary of the valid region closest to $X$
- A1: Correct real-world distance with units: 250 m (or 200 m with justified reasoning)
📋 Revision Sheet
| Term | Meaning |
|---|---|
| Locus | Set of all points satisfying a condition |
| Perpendicular bisector | Line at 90° through midpoint of a segment |
| Angle bisector | Line that divides an angle into two equal halves |
| Equidistant | At equal distances from two or more objects |
| Scale | Ratio of drawing length to real length ($1:n$) |
| Bearing | Clockwise angle from North, 3 digits |
Locus 1 (circle): $(x-a)^2 + (y-b)^2 = r^2$
Locus 2 (perp. bisector): $PA = PB$
Locus 3 (angle bisector): $d(P,\ell_1) = d(P,\ell_2)$
Locus 4 (stadium): Two parallel lines + two semicircles of radius $d$
Scale conversions: $\text{Real} = \text{Map} \times n$
Closer to A than B: $PA < PB$ → $A$-side of perp. bisector
- "P-PASS" — Point → circle; two Points → perp. bisector; Angle → angle bisector; Segment → Stadium
- "Arcs stay" — never rub out construction arcs
- "Same setting" — compasses must not be adjusted between pairs of arcs
- "BBC Bearing" — Bearing: Bearing, Clockwise from North, 3 digits
- "Locus vs Region" — "exactly" = line/curve; "within/more than" = shaded region
- "Two conditions → intersection" — shade the overlap, not either locus alone
- Always use compasses for bisectors — ruler-only attempts score zero
- Label every locus you draw (e.g., "perp. bisector of $AB$")
- State whether boundaries are included or excluded
- For scale drawings, write the scale used and show your conversions
- Check: perpendicular bisector should cross $AB$ at its midpoint at 90°
- In multi-constraint problems, construct each locus separately first, then identify the overlap
- Read "closer to A than to B" carefully — shade $A$'s side, not $B$'s
🔄 Flashcards
Click a card to reveal the answer. All 15 cards cover key facts and methods for this chapter.
✗ Common Mistakes
What students do: Rub out the compass arcs after drawing the construction line, thinking they look messy.
Why marks are lost: Method marks (B1) are awarded for visible arcs. Without them, the examiner cannot verify that compasses were used — you lose up to 2 marks on a 3-mark construction question.
How to avoid it: Never erase arcs. Draw them lightly so they're neat but visible. Write "construction arcs" next to them if unsure.
What students do: Change the compass setting between placing on $A$ and placing on $B$ when constructing a perpendicular bisector.
Why marks are lost: The arcs will not cross at the equidistant point, so the "bisector" will not be perpendicular or will not pass through the midpoint. The entire construction is invalid.
How to avoid it: Lock the compass setting after the first pair of arcs. Check it against a ruler before and after. The setting must be more than half of $AB$.
What students do: Draw only the two parallel lines for the locus "at distance $d$ from a line segment," omitting the semicircular ends.
Why marks are lost: The locus is incomplete. Points near the endpoints of the segment are still within distance $d$ — their locus traces the semicircles. Without them, the shaded region is wrong.
How to avoid it: Always ask: "what happens at the endpoints?" — the answer is always semicircles of radius $d$.
What students do: When told "closer to $A$ than to $B$," shade the $B$-side of the perpendicular bisector instead of the $A$-side.
Why marks are lost: The entire region is wrong. All subsequent work based on this incorrect shading loses marks.
How to avoid it: Pick a test point obviously close to $A$ (e.g., point $A$ itself) and check it satisfies the condition. If it does, shade that side. Always verify with a test point.
What students do: When asked for "all points within 3 cm of $P$," draw only the circle (the boundary) without shading the interior.
Why marks are lost: "Within" is an inequality ($<$ or $\leq$), so the answer is a region (disc), not just the boundary (circle). Marking scheme penalises not shading when asked for a region.
How to avoid it: "Exactly $r$ from" → draw the circle only. "Within $r$ of" → shade the interior too. "More than $r$ from" → shade the exterior.
What students do: Confuse which way to convert. For scale $1:200$, mistakenly divide a map distance by 200 to get the real distance (instead of multiplying), or apply the scale to the wrong measurement.
Why marks are lost: All distances on the diagram are wrong, making the entire locus construction invalid. A 1-mark conversion error cascades to lose all subsequent marks.
How to avoid it: Write the conversion as a sentence first: "1 cm on map = 200 cm in real life." Then: real distance = map distance × 200. Draw a small conversion table in the margin.
✅ Final Checklist
Click each item as you master it. Progress is saved automatically.
- I can describe the locus of points at a fixed distance $r$ from a point (circle, centre $P$, radius $r$)
- I can construct a perpendicular bisector accurately using compasses and leave arcs visible
- I can describe the locus equidistant from two fixed points (perpendicular bisector of $AB$)
- I can construct an angle bisector accurately using compasses
- I can describe the locus equidistant from two lines (angle bisector)
- I can draw the complete locus of points at a fixed distance from a line segment (parallel lines + semicircles)
- I can construct the perpendicular from an external point to a line
- I can identify and shade regions for compound locus problems with two or more constraints
- I understand the difference between a locus (boundary) and a region (inequality, shaded area)
- I can apply loci to scale drawings, converting between map and real distances correctly
- I can describe loci algebraically using equations and inequalities (e.g., $PA \leq r$, $PA < PB$)
- I can identify the intersection of loci as the solution to multi-constraint problems
- I know that arcs must be left visible on construction diagrams for method marks
- I can verify which side of a perpendicular bisector to shade using a test point
- I can tackle Grade 9 problems combining scale drawings, bearings, and multiple locus conditions