Secondary Mathematics · Worked Repair Guide 14
Construction and loci questions are not mainly drawing exercises. They are problems about conditions. A perpendicular bisector is the set of points equidistant from two endpoints. An angle bisector is the set of points equidistant from the two sides of an angle. A circle is the set of points at a fixed distance from a centre.
This guide develops one central habit: name the geometric condition before choosing the construction. Once the condition is clear, the ruler-and-compass action becomes a way of enforcing that condition rather than a memorised sequence of arcs.
All examples are original teaching constructions. Where a learner’s course uses physical instruments, accuracy and visible construction arcs matter.
1. Construction means creating a relationship exactly
A scale drawing may approximate a shape by measurement. A classical ruler-and-compass construction uses geometric relationships to locate a point or line exactly within the model.
For example, measuring the midpoint of a 10 cm segment at 5 cm uses length measurement. Constructing its perpendicular bisector finds the midpoint through equal-radius arcs from both endpoints.
Entry check: if point P lies on the perpendicular bisector of AB, then PA=PB. That equality is the defining condition.
Construction marks should therefore communicate why the result is valid, not merely where a final line happens to appear.
2. Constructing a perpendicular bisector
Given segment AB, choose a compass radius greater than half of AB. Draw arcs centred at A above and below the segment. Without changing the radius, draw arcs centred at B that intersect the first pair. Join the two arc-intersection points.
The resulting line is perpendicular to AB and passes through its midpoint.
Why? Each intersection point is the same compass radius from A and B, so it is equidistant from A and B. Two distinct equidistant points determine the perpendicular bisector.
Do not use two different compass radii from A and B. Equal radii are what preserve the equidistance condition.
3. Constructing a perpendicular through a point on a line
Let P lie on line l. Draw an arc centred at P to cut the line at A and B. Then construct the perpendicular bisector of AB. Since P is the midpoint of AB, that perpendicular bisector passes through P.
This construction reduces a new problem to a known one: create two points symmetric about P, then bisect the segment joining them.
The final line forms a 90° angle with l at P.
A set square may draw a practical perpendicular, but the compass construction shows the relationship explicitly when the question asks for a formal construction.
4. Constructing a perpendicular from a point outside a line
Let P lie outside line l. Draw an arc centred at P that cuts l at A and B. Since PA=PB, P lies on the perpendicular bisector of AB. Construct that perpendicular bisector; it passes through P and meets l at a right angle.
The geometry again comes from equidistance. The compass first manufactures two points on the line equally distant from P.
This is stronger than “drop a vertical-looking line”. A perpendicular is defined by angle, not by page orientation.
If the original line is slanted, the correct perpendicular will be slanted too.
5. Constructing an angle bisector
Given ∠ABC, centre the compass at B and draw an arc meeting BA and BC at P and Q. With equal radius, draw arcs centred at P and Q that intersect at R inside the angle. Join B to R.
BR bisects the angle, creating two equal angles.
The construction works because P and Q are equally distant from B, while R is equally distant from P and Q. The resulting triangles have matching side information that supports equality of the two angles at B.
The defining locus property is also useful: points on an angle bisector are equidistant from the two sides of the angle.
6. Constructing standard angles from familiar geometry
An equilateral triangle produces 60° angles. Constructing a perpendicular produces 90°. Bisecting 90° produces 45°. Bisecting 60° produces 30°.
These angle constructions are combinations of a small number of structural moves rather than isolated recipes.
Worked example: to construct 30° from a ray, first construct 60° using an equilateral-triangle arc relationship, then bisect the 60° angle.
When the question specifies ruler and compass, measuring 30° with a protractor does not demonstrate the requested construction method.
7. A locus is a set of all points satisfying a condition
The word locus is singular; loci is plural. A locus is not one chosen point. It is the complete set of positions satisfying a rule.
“Points 4 cm from A” form a circle centred at A with radius 4 cm.
“Points equidistant from A and B” form the perpendicular bisector of AB.
“Points equidistant from two intersecting lines” lie on their angle bisectors. Depending on the full setting, both internal and external angle bisectors may be relevant.
8. Distance from a point gives a circle
If a point P must be exactly 5 cm from A, then AP=5. Every point satisfying this forms the circumference of a circle with centre A and radius 5 cm.
If the condition is “no more than 5 cm from A”, the solution includes the disk inside and on the circle.
If the condition is “more than 5 cm from A”, the exterior region is required, excluding the boundary circle.
Words such as exactly, less than, at most and greater than determine whether the locus is a line, boundary or region.
9. Distance from a line gives parallel lines
The set of points exactly 3 cm from a straight line consists of two lines parallel to the original, one on each side, each at perpendicular distance 3 cm.
Distance from a line is measured perpendicularly, not diagonally.
“Within 3 cm of the line” describes the strip between those two parallels, including the original line.
A common mistake is to draw arcs around selected points on the line. The condition applies to every point along the straight line, so the boundary is parallel.
10. Equidistant from two points gives a perpendicular bisector
For fixed points A and B, the locus PA=PB is the perpendicular bisector of AB.
This fact turns many “find a possible location” problems into a construction problem.
Worked example: a sensor must be equally far from two gates A and B. Every possible position lies on the perpendicular bisector of AB. If a second condition is added, the valid sensor locations are intersections with that second locus.
One condition usually produces a whole locus; two independent conditions may narrow the answer to one or more intersection points.
11. Equidistant from two lines gives angle bisectors
For two intersecting lines, points with equal perpendicular distance from the lines lie on angle bisectors.
If the problem considers only the interior of one angle, the internal angle bisector is normally the relevant branch.
If the entire plane is considered, both bisector lines can satisfy the equal-distance condition.
Read the domain carefully. A diagram may show only one region even though the pure locus condition has more than one branch.
12. Combining loci means intersecting conditions
Suppose P must be 4 cm from A and equidistant from A and B. The first condition gives a circle centred at A radius 4. The second gives the perpendicular bisector of AB.
The valid points are where the circle intersects the perpendicular bisector.
There may be two, one or no intersections depending on the geometry.
This is the same logical structure as solving simultaneous equations: each condition defines a set, and the solution belongs to all sets at once.
13. Regions come from inequality-style geometric language
“Closer to A than B” means PA<PB. The perpendicular bisector of AB is the boundary where PA=PB.
The half-plane containing A satisfies PA<PB; the other half-plane satisfies PB<PA.
Test a point to identify the correct side rather than relying on visual intuition.
This connects loci to the Inequalities, Intervals and Regions guide: both topics convert conditions into sets and boundaries.
14. Bearings and loci can work together
Suppose B lies on a bearing 060° from A and exactly 8 cm away on a scale diagram. The bearing gives a ray; the distance gives a circle centred at A. Their intersection locates B.
A bearing condition alone gives infinitely many possible points along a ray. A distance condition alone gives infinitely many points on a circle. Together they can produce one point.
Draw the north line and measure clockwise if the task uses bearings. Then preserve the scale used in the diagram.
Use the Angles, Similarity and Geometric Reasoning guide for bearing conventions.
15. Scale drawings convert construction distance into real distance
If a plan uses scale 1:100, a constructed locus 3 cm from a wall represents an actual perpendicular distance of 300 cm=3 m.
Keep all construction distances in drawing units until the geometry is complete, then convert back to real units.
A scale factor changes length, while area and volume scale differently. Do not treat a 1:100 length scale as a 1:100 area ratio.
The Units, Scale and Measurement guide develops this further.
16. Capstone: locate a point satisfying three conditions
On an invented plan, point P must be equidistant from A and B, no more than 5 cm from A, and on the same side of AB as point C.
First construct the perpendicular bisector of AB. This is the equality condition PA=PB.
Next draw the circle centred at A radius 5 cm. “No more than 5 cm” means points inside or on the circle.
Finally restrict the perpendicular-bisector portion to the half-plane on the same side of AB as C.
The answer is a segment or ray portion of the perpendicular bisector lying inside the disk and in the required half-plane. A point may be chosen only from that remaining set.
17. Independent practice
- State the defining property of the perpendicular bisector of AB.
- State the defining property of an angle bisector in terms of angles.
- Describe the locus of points exactly 4 cm from A.
- Describe the locus of points at most 4 cm from A.
- Describe the locus of points exactly 2 cm from a straight line.
- Describe the locus of points equidistant from A and B.
- Describe the locus of points equidistant from two intersecting lines.
- Explain why equal compass radii are used from A and B in a perpendicular-bisector construction.
- What standard angle can be built from an equilateral triangle?
- How can 30° be constructed from 60°?
- A point is 6 cm from A and equidistant from A and B. What two loci should be drawn?
- What identifies the valid point or points in Question 11?
- Describe the region closer to A than B.
- At scale 1:200, what real distance does 4 cm represent?
- At scale 1:50, what drawing distance represents 6 m?
- Explain why a perpendicular is not necessarily vertical on the page.
- Explain why one locus condition usually does not determine one unique point.
- What happens if two loci do not intersect?
- What does “within 3 cm of a line” describe?
- Why should construction arcs usually remain visible when the method is being assessed?
18. Worked answers
1. Every point on it is equidistant from A and B.
2. It divides the angle into two equal angles.
3. A circle centred at A with radius 4 cm.
4. The disk inside and on that circle.
5. Two parallel lines, one on each side, at perpendicular distance 2 cm.
6. The perpendicular bisector of AB.
7. Their angle bisectors.
8. Equal radii make the arc intersections equidistant from A and B.
9. 60°.
10. Bisect the 60° angle.
11. A circle centred at A radius 6 cm and the perpendicular bisector of AB.
12. Their intersection points.
13. The half-plane on A’s side of the perpendicular bisector, excluding the boundary if “closer” is strict.
14. 8 m.
15. 12 cm. Six metres is 600 cm; divide by 50.
16. Perpendicularity is defined by a 90° angle to the given line, not by page orientation.
17. A single condition normally defines infinitely many positions.
18. No point satisfies both conditions simultaneously.
19. A strip between two lines parallel to the original, 3 cm away on each side.
20. They provide evidence that the requested geometric construction, rather than simple measurement, was used.
19. Diagnose the first geometric-condition error
Common failures include choosing the wrong locus, measuring distance diagonally from a line, forgetting the second branch of an equal-distance locus, shading the wrong region, using scale conversions in the wrong direction, or drawing a shape that looks correct without enforcing the defining condition.
A useful correction note names the condition: “equidistant from two points → perpendicular bisector”, “fixed distance from point → circle”, or “closer to A than B → A-side of the perpendicular-bisector boundary”.
Then change the diagram while keeping the same condition. Transfer is demonstrated when the learner can reconstruct the locus even after the page orientation and dimensions change.
20. Continue through the BTT Mathematics library
Return to the BTT Mathematics Hub. Use Angles, Similarity and Geometric Reasoning for angle facts and bearings, Units, Scale and Measurement for scale drawings, and Circle Geometry: Arcs, Sectors, Chords and Tangents for circle-based conditions.
The BTT Mathematical Lab is the diagnostic route when the same diagram-reading or construction-selection error keeps returning.
21. Sources and scope
The construction tasks, plan contexts and locus descriptions in this guide are original teaching material. They are designed to make geometric conditions explicit rather than reproduce examination questions.
For the current Singapore Secondary curriculum doorway, see MOE: Curriculum for secondary schools. Match formal constructions and loci extensions to the learner’s actual subject level and school programme.