1MA1 · G2, G15Geometry & MeasuresFoundation & HigherGrades 3–6

Constructions & Loci

Q: How accurate do constructions need to be?

Typically within 2 mm for lengths and 2^ for angles.

With only a ruler and a pair of compasses you can bisect an angle, draw a perpendicular exactly through the middle of a line, and mark out every point a given distance from a landmark. These $\textbf{constructions}$ solve $\textbf{loci}$ problems: where can a goat tethered to a post graze? Which part of a park is closer to the café than the gate?

45 min Angles & Geometric Reasoning

The big picture

Constructions work because compasses draw points at exactly equal distances. Every standard construction uses that fact — two arcs of the same radius crossing at points equidistant from two things — which is why the arcs must be left visible as evidence.

A locus is a set of points obeying a rule. Four loci cover almost every GCSE question: a fixed distance from a point (circle), a fixed distance from a line (parallel lines with semicircular ends), equidistant from two points (perpendicular bisector), and equidistant from two lines (angle bisector).

What you'll learn

Construct the perpendicular bisector of a line segment
Construct the bisector of an angle and a perpendicular from a point to a line
Construct triangles from given information
Solve loci problems, shading regions that satisfy several conditions

Perpendicular bisector

To bisect line segment $AB$ at right angles: set the compasses to more than half the length of $AB$ . Draw arcs from $A$ above and below the line, then, with the same radius, arcs from $B$ crossing them.

Join the two crossing points with a straight line. It passes through the midpoint of $AB$ at $90^{\circ}$ .

Every point on this line is the same distance from $A$ as from $B$ .

Tip — Keep the compass width fixed for both sets of arcs, and never rub out construction arcs — they are the evidence for the marks.

Angle bisector and perpendiculars

To bisect an angle at vertex $V$ : draw an arc centred at $V$ crossing both arms. From each crossing point, draw arcs of equal radius that meet inside the angle. Join $V$ to the meeting point.

Every point on the angle bisector is equidistant from the two arms.

To draw a perpendicular from a point $P$ to a line: draw an arc from $P$ cutting the line twice, then construct the perpendicular bisector of those two points; it passes through $P$ .

Why this works

The shortest distance from a point to a line is along the perpendicular. That is why "distance from a line" in loci problems always means the perpendicular distance.

Constructing triangles

Given three sides (SSS): draw one side accurately, then use compasses set to each of the other lengths from its ends; the arcs cross at the third vertex.

Given two sides and the included angle (SAS), or two angles and a side (ASA), use a protractor for the angles and a ruler for the sides.

Label vertices and leave all arcs visible.

Your turnSame method, new numbers — try it before moving on.

Can a triangle be constructed with sides of 3 cm, 4 cm and 8 cm? Type yes or no.

Your answer

Show worked solution

Solution

1The two shorter sides must add to more than the longest side.
2 $3 + 4 = 7$ , which is less than 8, so the arcs never meet: no.

Answer: no

Loci and regions

Points a fixed distance from a point form a $\textbf{circle}$ . Points a fixed distance from a line segment form a "running track" shape: two parallel lines joined by semicircles.

Points equidistant from two points lie on the $\textbf{perpendicular bisector}$ . Points equidistant from two intersecting lines lie on the $\textbf{angle bisector}$ .

"Closer to $A$ than $B$ " means one side of the perpendicular bisector; "less than 3 m from $P$ " means inside a circle. Combine conditions by finding where all regions overlap, and shade that region.

Example 1A garden is a rectangle

ABCD

with

AB = 10

m and

BC = 6

m, drawn to a scale of 1 cm to 1 m. A tree is to be planted closer to

AB

than to

AD

and less than 4 m from

A

. Describe how to find the region.

1"Closer to

AB

than to

AD

": construct the bisector of angle

DAB

; the region is on the

AB

side.

2"Less than 4 m from

A

": draw an arc of radius 4 cm centred on

A

; the region is inside it.

3Shade where both conditions hold, inside the rectangle.

AnswerThe region inside the arc of radius 4 cm about

A

, between

AB

and the angle bisector of

DAB

Your turnSame method, new numbers — try it before moving on.

A straight fence is 10 m long. Find the area of the region within 3 m of the fence (on both sides, including the rounded ends), to 1 decimal place.

Your answer

Show hint

The region is a rectangle with a semicircle at each end.

Show worked solution

Solution

1Rectangle: $10 \times 6 = 60$ m².
2Two semicircles make a circle of radius 3: $9\pi = 28.27$ m². Total $88.3$ m².

Answer: $88.3$ m²

Tip — Convert real distances with the scale before drawing. With 1 cm to 2 m, a 5 m radius becomes 2.5 cm.

Think like an examiner

Leave all construction arcs visible.
Use a sharp pencil and draw lines accurately to within 2 mm and angles within $2^{\circ}$ .
Label and shade the final region clearly, with a key if needed.

Loci

\text{fixed distance from a point: circle}

Point locus.

\text{equidistant from two points: perpendicular bisector}

Two points.

\text{equidistant from two lines: angle bisector}

Two lines.

Watch out for these

Changing compass width between the two arcs of a bisector.

Use the same radius for both sets of arcs.

Measuring the midpoint with a ruler instead of constructing.

A construction question requires the arcs.

Drawing a fixed distance from a line segment without rounded ends.

The ends are semicircles centred on the endpoints.

Forgetting to apply the scale.

Convert real distances to map distances first.

Stretch yourself

Two phone masts $P$ and $Q$ are 8 km apart. A signal is received anywhere within 5 km of either mast. Explain which points on the perpendicular bisector of $PQ$ receive a signal, and find the length of that section.

Hint — Points on the perpendicular bisector are equidistant from $P$ and $Q$ . Use Pythagoras with half of $PQ$ .

Reveal solution

SolutionA point on the perpendicular bisector is the same distance

d

from both masts, so it receives a signal if

d \le 5

km. The bisector crosses

PQ

at its midpoint

M

, 4 km from each mast. A point on the bisector at height

h

above

M

\sqrt{16 + h^{2}}

km from each mast. So we need

16 + h^{2} \le 25

, giving

h \le 3

. The section of the bisector with a signal extends 3 km either side of

M

, a total length of 6 km.

\blacksquare

Questions students ask

What equipment is allowed?

A ruler, compasses and pencil for constructions; a protractor for angle-based triangles and bearings.

What is the plural of locus?

Loci.

How accurate do constructions need to be?

Typically within 2 mm for lengths and

2^{\circ}

for angles.

Key takeaways

Constructions rely on equal compass distances; leave arcs visible.
Perpendicular bisector: equidistant from two points.
Angle bisector: equidistant from two lines.
Fixed distance from a point is a circle; from a segment, a running-track shape.
Combine loci and shade the region satisfying all conditions.

Exam-style questions

Practise it the way it is examined

3 original questions · 7 marks, written to match the style and mark allocation of the real papers. Work on paper, then open the mark scheme and tick the marks you earned — method marks count even if the final answer slips.

Question 1Grade 31 mark

Describe the locus of points that are the same distance from two straight lines that meet at a point.[1]
Show mark scheme
- B1The bisector of the angle between the lines

Question 2Grade 53 marks

A goat is tied to a corner post of a large rectangular field by a rope 7 m long.

Work out the area of the field the goat can reach. Give your answer to 1 decimal place.[3]
Show mark scheme
- M1Recognises a quarter circle of radius 7 m
- M1 $\frac{1}{4} \times \pi \times 7^2$
- A1 $38.5$ m²
Answer: $38.5$ m²

Question 3Grade 63 marks

Describe how to construct, using only a ruler and compasses, the perpendicular from a point $P$ to a line $L$ .[3]
Show mark scheme
- B1Draw an arc centred on $P$ cutting $L$ at two points
- B1From each point, draw arcs of equal radius that intersect on the far side of $L$
- B1Join $P$ to the intersection

Test yourself

Answer a few questions, then review anything you missed.

Next step

Keep practising

Start with the quiz above to check the key ideas. You can also choose another topic from the practice library.

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