Pass an Exam 
Hi, I’m Ari. We can start constructions and loci from the beginning, work through an example together, or practise a question. Tell me which step is confusing.
GCSE Maths · Geometry & measures
Constructions and loci GCSE Questions and Worked Answers
A locus is the set of all points satisfying a condition, such as being a fixed distance from a point. Use circles for fixed point-distance, perpendicular bisectors for equal distance from two points, and angle bisectors for equal perpendicular distance from intersecting lines.
Free AI tutor · GCSE constructions and loci
Practise GCSE constructions and loci for free with an AI tutor
Ask Ari for an explanation or work through an original exam-style question together.
AriYour maths coach
Chat cost ≈ $0.000000
Ari is an AI tutor and can make mistakes. Use the worked answers below to check important results.
Start with the meaning
What you need to know about constructions and loci
Tie a pencil to a fixed point with a taut string. As the pencil moves, it stays the same distance from the fixed point and traces a circle. We call all the positions that satisfy a condition a locus; the plural is loci. Here the condition is 'exactly this distance from the point'. A compass draws that circle by keeping its opening fixed.
See the idea first
Equal compass radii create equal distances
Now ask for points equally far from two fixed points A and B. Draw equal-radius circles centred at A and B, with radius greater than half AB. Each intersection is the same distance from A as from B. The straight line through the intersections passes through the middle of AB at 90°: the perpendicular bisector. Every point on that line is equally far from A and B.
From problem to method
Typical problems you need to be able to solve
These are the main kinds of question in this topic. Each card first states the problem, then explains how to solve it.
A right angle at a point on a line
What the problem asks: P lies on a straight line. Construct a perpendicular through P without a protractor.
How to solve it: Mark A and B on the line with an arc centred on P, so PA = PB. Draw equal-radius arcs from A and B to meet above the line at Q, using a radius greater than PA. Join PQ. The matching triangles on either side of PQ make the two angles at P equal, so both are 90°.
Fixed distance from a point
What the problem asks: Draw all points exactly 4 cm from P.
How to solve it: Set the compass opening to 4 cm and draw a circle centred on P. Use the circumference only, not the interior.
Equal distance from two points
What the problem asks: Construct all points equally far from A and B.
How to solve it: Construct the perpendicular bisector using equal-radius arcs from A and B. Keep the arcs visible; the line, not just the arc intersections, is the locus.
Equal distance from two meeting lines
What the problem asks: Inside an angle, locate points equally far from its two sides.
How to solve it: Construct the internal angle bisector. Distance to a line is measured at 90°, so points on this bisector have matching perpendicular distances.
Meet two conditions
What the problem asks: Shade points within 3 cm of P and closer to A than B.
How to solve it: Draw the circle centred on P and the perpendicular bisector of AB. Keep only the part inside the circle on A's side of the bisector.
A reliable routine
Choose a locus before constructing it
Use this for positions constrained by distances. A compass encodes fixed distances, and bisectors encode equal distances. Translate each condition into its correct geometric set before combining them.
- Read exactly, within, at least, closer and equally carefully.
- Choose the circle, bisector or parallel boundary justified by the condition.
- Construct using ruler and compass, retaining the evidence arcs.
- For several conditions, keep only their intersection; state whether boundaries are included.
Check: A perpendicular bisector cuts a segment in half at 90°. An angle bisector splits an angle into two equal angles. A ruler-and-compass construction is not a protractor measurement.
Fully worked
Constructions and loci GCSE worked examples
Each solution explains the clue, why the method fits and how to check the result.
Example 1
A perpendicular at a point on the line
Question
Draw a straight line and mark P on it. Using ruler and compass only, construct a perpendicular at P. Explain why it makes a right angle.
Use a compass centred at P to mark A and B equally far along the line on either side. Increase the opening to more than PA. With this same opening from A and B, draw arcs meeting at Q off the line. Join PQ. Triangles APQ and BPQ have PA = PB, AQ = BQ and shared PQ. Their matching angles at P are equal and sum to 180°, so each is 90°. Tip: keep both sets of arcs; simply drawing a line that looks vertical is not a construction.
Construct a perpendicular at a point on a line
The task is to draw a line through P making a right angle with the original line. First create two points A and B equally far from P. Their perpendicular bisector will pass through P because P is their midpoint.
Why it works: PA = PB, AQ = BQ, and PQ is shared. The triangles APQ and BPQ have three matching sides, so the angles at P are equal. Together those angles make the 180° straight line; each is therefore 90°.
Example 2
Perpendicular bisector
Question
Draw AB = 6 cm. Construct its perpendicular bisector using ruler and compass. Explain the construction.
Draw AB. Set the compass radius to 4 cm, which is greater than 3 cm (half AB). Draw arcs above and below AB from A; without changing the radius, repeat from B. Join the two intersections with a straight line. Each intersection is 4 cm from both A and B, so their joining line is the perpendicular bisector. Leave the arcs visible.
Example 3
Bisect an angle
Question
Construct the internal bisector of an angle with vertex O.
With the compass centred at O, draw one arc crossing the two sides at P and Q. Then OP = OQ. Choose a radius greater than half PQ. Using that same radius from P and Q, draw intersecting arcs inside the angle at R. Join O to R. Triangles OPR and OQR have OP = OQ, PR = QR and shared OR, so SSS gives equal angles POR and QOR. Leave all arcs visible.
Example 4
A perpendicular from a point
Question
Construct a perpendicular from point P above a straight line l to l.
Draw an arc centred at P that crosses l at two points A and B, so PA = PB. With equal-radius arcs centred at A and B, construct another point Q off l that is equally far from A and B. Draw PQ. Both P and Q lie on the perpendicular bisector of AB, so PQ meets l at 90°. Choose Q distinct from P, typically on the other side of l, and retain the arcs.
Example 5
Exactly versus within
Question
Describe the loci of points (a) exactly 5 cm from A and (b) at most 5 cm from A.
(a) The circumference of a circle centred at A, radius 5 cm. (b) The whole disk inside that circle, including its circumference. 'At most' means distance ≤ 5 cm; it does not restrict points to the edge.
Example 6
Two fixed distances
Question
A = (0, 0) and B = (6, 0). Find points exactly 5 units from both A and B.
Equal distances put the points on the perpendicular bisector x = 3. A right triangle then has hypotenuse 5 and horizontal leg 3:
The two points are (3, 4) and (3, −4), the intersections of the two radius-5 circles.
Example 7
Combine a region
Question
A = (0, 0), B = (6, 0). Describe points at most 4 units from A and strictly closer to A than B.
The first condition is the disk centred at A with radius 4, including its circle. The equal-distance boundary is x = 3. Closer to A means x < 3. Keep the overlap: the part of the disk left of x = 3. Exclude the straight bisector boundary because 'strictly closer' does not allow equal distances.
11 original questions · total 21 marks
Constructions and loci GCSE exam-style questions
Try each question before opening its fully worked answer. The difficulty rises through the set.
Before you startAllow about 26 minutes · Show your working. Give exact answers unless a question asks you to round, and include units where needed. · answers start collapsed
Construct at the midpoint
Draw AB = 8 cm and mark its midpoint P. Construct the perpendicular through P with ruler and compass, leaving your arcs visible.
Show worked answer
P is 4 cm from each endpoint. Choose a compass radius greater than 4 cm, for example 5 cm. Draw arcs with this same radius from A and B to intersect at Q above the line. Join PQ. Because Q is equally far from A and B, it lies on their perpendicular bisector; that bisector passes through midpoint P. Leave the arcs visible.
Circle locus
Describe points exactly 2 cm from P.
Show worked answer
A circle centred at P with radius 2 cm, meaning its circumference only.
Two points
Describe all points equally far from A and B.
Show worked answer
The perpendicular bisector of AB: the line through its midpoint at 90°.
Two lines
Inside a 60° angle, what is the locus equally far from its two sides? What angle does it make with either side?
Show worked answer
The internal angle bisector.
It makes 30° with either side. Distances to the sides are measured perpendicularly.
Compass opening
AB is 8 cm. Why do equal arcs of radius 3 cm fail to construct its perpendicular bisector?
Show worked answer
The two circles do not meet: their radii add to 6 cm, less than AB = 8 cm. Use the same radius greater than 4 cm from each endpoint to obtain two intersections.
Construction evidence
Why should construction arcs remain visible?
Show worked answer
They show how the exact equal-distance or equal-angle condition was constructed, rather than guessed or measured only.
Distance from a line
Describe all points exactly 2 cm from an infinite straight line.
Show worked answer
Two parallel lines, one on each side, each at perpendicular distance 2 cm. It is not a single line unless a side is specified.
Closer to one point
A = (0, 0), B = (10, 0). Describe points strictly closer to B than A.
Show worked answer
The equal-distance boundary is x = 5. B lies to its right, so the required region is x > 5; exclude the boundary.
An annulus
Describe points at least 2 cm and at most 5 cm from P.
Show worked answer
The ring between concentric circles centred at P with radii 2 and 5 cm, including both boundaries. Distances must satisfy both limits.
Construct 60 degrees
Use ruler and compass to construct a 60° angle at A on segment AB = 5 cm.
Show worked answer
Draw circles or arcs of radius 5 cm centred at A and B. Choose an intersection C and join AC and BC. AB = AC = BC = 5 cm, so ABC is equilateral.
Thus angle CAB is 60°.
Count intersections
A and B are 10 cm apart. Can a point be exactly 3 cm from both? Explain.
Show worked answer
No. Two such radii would total only 6 cm, less than the 10 cm between the centres, so the circles cannot meet. Equivalently a triangle with sides 3, 3, 10 cannot exist.
Examiner-style feedback
Common constructions and loci mistakes
Equally far from two points gives a whole perpendicular-bisector line, not just the midpoint.
Distance from a point to a line is the shortest, perpendicular distance.
When both conditions must hold, keep their overlap, not every point satisfying either one.
30-SECOND RECAP
Choose, carry out, check
Translate the wording before you reach for a rule.
- Start with the distance condition.
- Choose the justified locus.
- Construct with visible arcs.
- Combine conditions and check boundaries.
Quick answers
Constructions and loci FAQ
Can I use a protractor to bisect an angle?
For a ruler-and-compass construction task, use compass arcs and a straightedge. Measuring half the angle does not demonstrate the required construction.
Does 'within' include the boundary?
Follow the exact wording. 'At most' includes it; 'less than' excludes it. If 'within' is ambiguous, make your boundary convention explicit.
Content standards
Curriculum and rights review
G2 ruler-and-compass constructions and loci across tiers. Drawings are explanatory; students should use physical instruments and retain arcs for construction practice. Questions, suggested marks, solutions and diagrams are original practice material, not official past-paper questions.
Official specification references