Loci Problems and Geometric Constructions Revision Notes | GCSE AQA Higher Maths

1. Why Constructions and Loci Matter

Imagine a mobile phone mast that must be placed so it is equidistant from two towns, or a security fence that must stay within 50 m of a building's boundary. These are real loci problems — and the tool you use to solve them is a pair of compasses and a ruler.

At Higher GCSE, you are expected to construct these diagrams accurately and use them to solve multi-constraint problems. The key word is construct: every arc you draw with compasses must remain visible on your final answer. Erasing arcs costs marks.

2. The Four Core Constructions

There are four constructions you need to have at your fingertips. Each one produces a specific locus, and knowing which construction matches which locus condition is half the battle in an exam question.

• Perpendicular bisector of a line segment — locus of points equidistant from two endpoints
• Angle bisector of an angle — locus of points equidistant from two lines
• Perpendicular from a point to a line — gives the shortest (perpendicular) distance from that point to the line
• Perpendicular at a point on a line — constructs a 90° angle at a specific point on a line

2a. Perpendicular Bisector of a Line Segment

The perpendicular bisector is the most frequently tested construction. It answers the question: which points are exactly the same distance from A as they are from B? The answer is every point on the perpendicular bisector of AB.

The resulting line passes through the midpoint of AB at exactly 90°. Every point on this line is equidistant from A and B — that is the locus condition it represents.

2b. Angle Bisector

The angle bisector answers a different question: which points are equidistant from both arms of an angle? Think of it as the fair dividing line between two directions.

2c. Perpendicular from a Point to a Line

This construction is the geometric proof of a key fact: the shortest route from a point to a line is always the perpendicular route. You will need this both as a construction and as a reasoning statement in exam questions.

The distance from P to the foot of this perpendicular is the perpendicular distance from P to L. This is always the shortest distance from P to the line — any other route from P to a point on L would be longer because it forms the hypotenuse of a right-angled triangle.

In the diagram, the vertical line from P to L is the perpendicular distance. The slanted line to any other point on L is the hypotenuse of a right-angled triangle — and the hypotenuse is always longer than either leg. That is why the perpendicular distance is always the shortest.

3. Standard Loci and Their Constructions

Now that the four constructions are in place, it is worth mapping each one to its locus condition. Exam questions describe a condition in words — your job is to recognise which construction to use.

4. Worked Example: Single-Constraint Locus

Single-constraint loci questions ask you to draw one locus. They test whether you can translate a word condition into the correct construction. Here is a typical exam-style question.

Worked example: Constructing the Locus of Points Equidistant from Two Points

Points A and B are 8 cm apart. Construct the locus of all points that are equidistant from A and B. [2 marks]

1. Equidistant from two points A and B $\Rightarrow$ perpendicular bisector of AB — Recognising the condition earns the first method mark [M1]. The locus is the perpendicular bisector of the line segment AB.
2. Set compass to a radius greater than $\frac{8}{2} = 4$ cm, e.g. 5 cm — The radius must exceed half of AB so the arcs from A and B actually intersect. If they do not intersect, no bisector can be drawn.
3. Place compass point on A. Draw an arc above and below line AB. — Keep the compass radius fixed throughout this construction.
4. Without changing the radius, place compass point on B. Draw arcs that cross the arcs from A at two points — call them P and Q. — Using the same radius ensures P and Q are equidistant from both A and B.
5. Use a ruler to draw a straight line through P and Q. — This line is the perpendicular bisector of AB. It passes through the midpoint of AB at 90°. [A1] for correct construction with arcs visible.

Answer: The perpendicular bisector of AB, drawn through the two arc intersections, with all construction arcs left visible.

5. Multi-Constraint Loci Problems

This is where Higher loci problems get genuinely interesting. Instead of one condition, you are given two or more — and the answer is the region or point where all conditions are satisfied simultaneously.

The strategy is always the same: construct each locus separately, then identify the overlap. Shading the correct region (or marking the correct intersection point) is what earns the final accuracy marks.

Worked example: Multi-Constraint Locus: Region Satisfying Two Conditions

A rectangular field ABCD has AB = 10 cm and BC = 6 cm on a scale diagram. A tree T is to be planted so that: (i) T is closer to A than to B, and (ii) T is more than 3 cm from the line BC. Construct and shade the region where T can be planted. [4 marks]

1. Closer to A than to B $\Rightarrow$ on the A-side of the perpendicular bisector of AB — The perpendicular bisector of AB divides the plane into two halves. Points closer to A lie on A's side. [M1] for identifying this construction.
2. Set compass radius $> 5$ cm. Draw arcs from A and B. Join intersections to form the perpendicular bisector of AB. — Leave all arcs visible. The bisector passes through the midpoint of AB at 90°. [M1] for correct construction with arcs.
3. More than 3 cm from line BC $\Rightarrow$ on the far side of a line parallel to BC, drawn 3 cm away from BC (into the field) — The locus of points exactly 3 cm from a straight line is a parallel line. Points more than 3 cm away are on the far side of this parallel line.
4. Measure 3 cm from BC towards AD. Draw a line parallel to BC at this distance. — This boundary line represents the minimum distance from BC. [M1] for correct parallel line.
5. The required region is: on the A-side of the perpendicular bisector of AB AND more than 3 cm from BC (i.e., beyond the parallel line, away from BC). — Shade the area inside the field that satisfies both conditions simultaneously. [A1] for correct region shaded.

Answer: The shaded region is bounded by the perpendicular bisector of AB (on A's side) and the line parallel to BC at 3 cm distance (on the far side from BC), within the rectangle ABCD.

Notice how each condition was handled separately before combining them. This two-stage approach prevents confusion and ensures you earn method marks even if the final shading is slightly off.

6. The Perpendicular Distance Principle

The specification requires you to know that the perpendicular distance from a point to a line is the shortest distance to that line. This fact underpins several loci problems and can appear as a reasoning question in its own right.

Here is the reasoning you need to be able to reproduce. Suppose P is a point not on line L, and F is the foot of the perpendicular from P to L. For any other point Q on L, triangle PFQ is right-angled at F. By Pythagoras' theorem, $PQ^2 = PF^2 + FQ^2$. Since $FQ^2 > 0$, it follows that $PQ > PF$. So PF — the perpendicular distance — is strictly less than any other distance from P to L.

This principle matters in loci problems because whenever a question says "the shortest distance from point P to line L", you must draw the perpendicular from P to L — not any other line.

Worked example: Finding the Shortest Distance from a Point to a Line

On a scale diagram, point P is 5 cm from line L (measured perpendicularly). A student claims the distance from P to a point Q on L, where PQ is at 60° to L, is shorter. Show that this claim is incorrect. [3 marks]

1. $PF = 5$ cm, where F is the foot of the perpendicular from P to L. — PF is the perpendicular distance — the candidate shortest distance. [M1] for identifying PF correctly.
2. In right-angled triangle PFQ, the angle at Q between $PQ$ and line $L$ is $60^\circ$. Because $PF$ is perpendicular to $L$, the angle at F is $90^\circ$, so the remaining angle at P is $30^\circ$. Using $\sin 30^\circ = \frac{PF}{PQ}$ gives $PQ = \frac{PF}{\sin 30^\circ} = \frac{5}{0.5} = 10$ cm. — The 30° angle is at P, not at F. F is the right angle because PF is perpendicular to line L. [M1] for correct trigonometric setup.
3. $PQ = 10$ cm $> PF = 5$ cm — PQ is twice the perpendicular distance. The oblique path is longer, not shorter. [A1] for correct comparison and conclusion.
4. Since $PQ > PF$, the perpendicular distance PF = 5 cm is shorter than PQ = 10 cm. The student's claim is incorrect. — Always state the conclusion explicitly in a 'show that' or reasoning question — the examiner needs to see you have addressed the claim directly.

Answer: PQ = 10 cm > PF = 5 cm, so the perpendicular distance is shorter. The student's claim is incorrect.

7. Exam Technique and Common Mistakes

Construction and loci questions follow predictable patterns, and so do the mistakes that cost marks. The guidance below is drawn directly from the types of errors that appear in mark scheme reports for Higher-tier papers.

8. Practice Problems

These problems cover the full range of construction and loci skills from this lesson. Work through each one before checking the answer and solution.

Every construction and loci problem at Higher GCSE reduces to one or more of these five cases. Recognise the condition in the question, select the matching construction, and combine loci where multiple conditions are given — that is the complete method for this topic.