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aiMOOC

Geometric constructions and loci - Mathematics Year 9

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Introduction

Why do equal compass arcs locate a midpoint or divide an angle equally? Build the two constructions, keep the arcs visible, and connect their steps to equal-distance reasoning.

Produce a perpendicular bisector and an angle bisector with labelled evidence. A simple locus is an editorial extension here, not an additional official DfE Year 9 requirement. Online reasoning supports practical work; actual instrument accuracy needs a real construction to inspect.

Construct and justify a perpendicular bisector

A perpendicular bisector crosses a segment at its midpoint and at 90°. The word “bisector” supplies equal halves; “perpendicular” supplies the right angle. A line meeting AB at 90° somewhere other than its midpoint satisfies only one part of that definition.

First practical construction.

Step	Action	Evidence to retain
1	Draw segment AB	Keep distinct endpoints and the given length
2	Choose compass radius r greater than half AB	Two equal circles must meet in two points
3	Draw arcs from A above and below AB	Retain the chosen opening
4	Use the same radius from B	Mark common arc intersections U and V
5	Join U, V with a straightedge	Call its intersection with AB midpoint M

The common radius must be greater than half the distance AB. If it is smaller, the equal circles do not meet; if exactly half, they only touch at the midpoint and do not supply two distinct points for a construction line. It need not equal the whole segment length. Do not change the opening between centres A and B.

Each intersection U is the same distance from A and B because both drawn arcs have the same radius; the same holds for V. Joining the two distinct intersections gives the perpendicular bisector. Keeping the arcs visible lets another person check how these points were obtained rather than seeing only a final line that looks central.

Original coordinate check. Set A=(−3, 0), B=(3, 0), and use radius 5 units. U=(0, 4) and V=(0, −4) lie 5 from each endpoint because 3²+4²=25. Their common line is x=0, which crosses AB at M=(0, 0), three units from either endpoint. The vertical and horizontal lines are perpendicular. These are defined mathematical coordinates, not measurements read from the reference image.

For a general point P=(x, y), equality of squared endpoint distances gives $(x + 3)^{2} + y^{2} = (x - 3)^{2} + y^{2}$ . Expanding and cancelling yields 12 x=0, so x=0. Conversely every point(0, y) has equal squared distances 9+y² to A and B. This proves both directions of the equal-distance property in this coordinate example, not just that two selected arc intersections work.

A ruler-and-compass method transfers to a rotated segment without needing coordinate axes. Coordinate arithmetic gives an independent check for chosen coordinates, while actual compass use requires careful openings and retained intersections. Measuring the midpoint first and drawing a right angle with a protractor is another drawing method, but it is not evidence that the requested compass construction was completed.

Construct an angle bisector with equal-arc evidence

Animation of equal compass circles around endpoints A and B,followed by their common perpendicular-bisector line. — The animation supplies a repeatable construction sequence. Pause or revisit the written steps as needed;the equal-radius and two-intersection conditions are stated in text rather than conveyed only by motion.

An angle bisector divides an angle into two equal angles from the same vertex. It is a ray within the stated non-reflex angle, not any line that appears to pass through a visually central point. Keep the vertex and both original arms identifiable.

Second practical construction. For angle AOB, start with centre O and draw one arc crossing the two arms at A and B. This makes OA=OB. From A and B, draw arcs of the same second radius large enough to intersect inside the angle at a point W other than O. Join O to W and extend the ray inside the angle. Retain both sets of compass arcs and label the constructed ray.

The justification compares triangles OAW and OBW. OA=OB from the first arc, AW=BW from the second pair, and OW is shared. SSS congruence makes angles AOW and WOB equal. Equal-looking halves or a protractor reading may check the result approximately, but the equal-radius facts explain the exact construction.

Original right-angle task. Let O=(0, 0), A=(3, 0), B=(0, 3). The angle AOB is 90°. Circles centred at A and B with radius 3 meet at O and W=(3, 3); choose W rather than the vertex itself. The ray from O through W lies on y=x and makes 45° with each arm. Check OA=OB=3, AW=BW=3, and OW²=18. Both triangles have the same three side lengths.

This numerical task illustrates a 90° case; the equal-arc SSS reasoning applies to other ordinary non-reflex angles too. Do not assume every angle bisector creates 45° halves. For a stated 74° angle, the halves are 37°, while a 120° angle gives 60°. A bisector halves the given angle, not a memorised right angle.

Watch the original Corbettmaths perpendicular-bisector demonstration and identify the equal-radius arcs before its final line appears. Its visible segment in a quadrilateral is separate from our A=(−3, 0), B=(3, 0) task; the angle-bisector task uses the distinct written procedure above. Original: corbettmaths, Perpendicular Bisectors - Corbettmaths. The written explanations and native tasks work without sound or the external player.

Compare both constructions. The segment task uses two endpoint centres to locate points equally far from the endpoints. The angle task first creates equal distances from its vertex, then uses matched endpoint arcs and SSS. Mixing their centres or silently changing an opening breaks the reason the final line is justified.

Retain two actual labelled drawings with construction arcs. A typed account of the steps is useful conceptual evidence, but it does not demonstrate physical instrument accuracy. A teacher or peer can inspect the actual work against the retained conditions and approximate measurement checks.

Interpret a locus and test all-points conditions

A locus is the set of all points satisfying a stated condition. “All” matters: showing that one or two points work does not establish the entire set. Here locus language is an editorial interpretation of familiar distance constructions, not a separate statutory Year 9 allocation.

For distinct endpoints A, B in a plane, the locus of points P with $P A = P B$ is the whole perpendicular bisector of AB. It extends in both directions beyond the segment and its midpoint. It is not merely the midpoint and is not the segment AB itself. The coordinate derivation in chapter 1 shows why every point of x=0 works and why no other x does for A=(−3, 0), B=(3, 0).

Check membership using squared distances. Both distances are nonnegative, so comparing their squares preserves equality and order without a rounded square root.

Defined point	Distance squared to A	Distance squared to B	Interpretation
P=(0, 2)	13	13	On the equidistance line
Q=(1, 2)	20	8	Closer to B; not equidistant
M=(0, 0)	9	9	Midpoint and on the line

Point P gives 3²+2²=13 from each endpoint. Point Q gives 4²+2²=20 from A and(−2)²+2²=8 from B, so it is closer to B. A midpoint is one member of the locus, but the first row shows another member away from AB. “P is nearer A than B” would instead describe one open half-plane; it would exclude the equal-distance boundary.

Another elementary locus is the set exactly 4 cm from a fixed point O: it is the circumference of a circle of radius 4 cm. The centre is not a member and the interior is not included. “At most 4 cm” includes the interior and boundary as a closed disc; “less than 4 cm” excludes the circumference. Keep exact equality distinct from a range condition.

Unfamiliar transfer. Set A=(1, 2), B=(7, 2). Their midpoint is(4, 2), so the perpendicular bisector is x=4. At P=(4, 6), both squared distances are 3²+4²=25. A common compass radius 5 gives intersections(4, 6) and(4, −2). This new segment changes the position of the same construction; the equal-distance reasoning remains valid.

First-error analysis. If endpoint arcs use different radii, their intersections need not be equidistant from A and B. The first error is the changed opening, not necessarily the final line's visible tilt. If the common radius equals half AB, one touching point cannot determine the required construction line. If an angle's first arc has different radii on its two arms, OA=OB is no longer a justified premise.

Retain both constructions, one all-points locus explanation, membership checks and a new-value transfer. Keep exact geometric reasoning separate from finite-precision drawing checks. Record actual later retrieval and actual instrument work when available; do not replace missing practical evidence with a claimed perfect construction.

Interactive Tasks

Quiz: Test Your Knowledge

A perpendicular bisector must cross AB where? (At its midpoint and at 90°) (!Anywhere at 90°) (!At endpoint A only) (!At its midpoint at any angle)

For two distinct equal-circle intersections, the common radius must be? (Greater than half AB) (!Less than half AB) (!Zero) (!Exactly half AB)

Why is a common arc intersection equally far from A and B? (Both circles use the same radius) (!The drawing looks symmetric) (!Every point in a circle is a centre) (!A ruler automatically proves equality)

In the coordinate task, A −3, 0 andB 3, 0 have midpoint? (0, 0) (!3, 3) (!−3, 3) (!6, 0)

Angle-bisector proof uses which triangle criterion here? (SSS) (!AAA alone) (!SSA in every case) (!A visual guess)

A 74° angle bisects into? (37° each) (!45° each) (!74° each) (!148° each)

The locus equally far from two distinct points is? (Their whole perpendicular bisector) (!Only their midpoint) (!The whole plane) (!Their connecting segment)

Exactly 4 cm from O describes? (A circle circumference) (!Only the centre) (!The entire disc including centre) (!All points beyond 4 cm)

At Q 1, 2 in the endpoint task, which squared distances occur? (20 and 8) (!13 and 13) (!9 and 9) (!8 and 20)

New endpoints 1, 2 and 7, 2 have bisector? (x=4) (!y=4) (!x=2) (!y=7)

Memory Game

Perpendicular	Meeting at a right angle
Bisector	Divider into two equal parts
Compass radius	Opening defining distance from a centre
Locus	Set of all points satisfying a condition
Midpoint	Point halfway between segment endpoints
Congruence	Same shape and size supporting equal angles

Drag and Drop

x=0	Original endpoint equidistance line
0, 4 and 0, −4	Original radius 5 arc intersections
SSS	Angle-bisector triangle justification
37°	Each half of the stated 74° angle
x=4	New translated-segment equidistance line

Crossword Puzzle

BISECTOR	Divider into two equal parts
PERPENDICULAR	At right angles
LOCUS	All points satisfying a condition
MIDPOINT	Halfway point of a segment
RADIUS	Distance defining a circle from its centre
INTERSECTION	Point belonging to both construction arcs

LearningApps

Optional extra practice. Choose an appropriate activity with your teacher. If the external collection is unavailable, use the native activities and independent tasks in this course.

Cloze Text

Open Tasks

Easy

Identify two required properties: State midpoint and 90° for a perpendicular bisector.
Check original arc intersections: Use 3²+4²=25 for U and V.
Halve two given angles: Calculate 74°→37° and 120°→60°.
Read an exact-distance condition: Distinguish a circumference from a disc.

Standard

Produce the first construction: Keep distinct endpoints, equal-radius arcs, U, V and midpoint M.
Produce the second construction: Keep the vertex arc, matched endpoint arcs, inner point W and ray OW.
Explain the full equidistance locus: Give both directions using the original coordinate condition.
Check locus membership: Use squared distances for P, Q and M.

Advanced

Transfer the construction: Use endpoints 1, 2 and 7, 2 with radius 5 and check both intersections.
Prove angle equality: Match OAW and OBW by SSS and infer equal vertex angles.
Locate a first construction error: Discuss unequal radii or a tangent-circle radius equal to half AB.
Retain practical evidence honestly: Keep both actual drawings, a measured check and real later retrieval separately from conceptual answers.

Learning Check

First accurate construction: Use a common radius above half AB and retain both arc intersections.
Second accurate construction: Show both equal-radius stages and the angle-bisector proof.
All-points locus reasoning: Explain equality and its converse, not just two sample points.
Membership and boundary checks: Interpret squared distances and fixed-distance sets.
Unfamiliar transfer and first error: Check the moved segment and repair a broken arc condition.
Portfolio and practical limits: Retain both constructions, the locus argument and actual evidence.

Worked checks and assessment guidance

Original A −3, 0 and B 3, 0 with radius 5 give U 0, 4 and V 0, −4, since all four endpoint distances square to 25. Line x 0 meets AB at midpoint 0, 0 and at 90°. Inspect actual arcs and drawing accuracy separately.
OA=OB from the vertex arc; AW=BW from endpoint arcs; OW shared. OAW≅OBW by SSS, so vertex angles equal. In the right-angle example W 3, 3 gives 45° halves; 74° instead gives 37° halves.
For A −3, 0 and B 3, 0, equality(x+3)²+y²=(x −3)²+y² reduces 12 x 0, so x 0. Conversely every 0, y gives matching 9+y². The whole perpendicular bisector is included, not only the midpoint.
P 0, 2 gives 13/13; Q 1, 2 gives 20/8 and is closer B; midpoint 0, 0 gives 9/9. Exactly 4 cm from O is the circumference; at most 4 cm is the closed disc.
Endpoints 1, 2 and 7, 2 give midpoint 4, 2, bisector x 4, radius 5 intersections 4, 6 and 4, −2, all squared distances 25. Unequal radii remove equidistance; radius exactly half AB gives only one touching point.
Keep two labelled compass-and-straightedge drawings with arcs, reasoning, a first-error correction and transfer. Coordinate checks support exact geometry but do not certify unobserved instrument competence; record actual inspection and later retrieval when performed.

Learning evidence

Retain two actual labelled constructions with visible arcs, a full equal-distance locus explanation, membership checks, first-error correction and unfamiliar transfer. Practical accuracy requires the actual work to be inspected.

OER and Wikipedia

Department for Education: England mathematics programme of study, Key Stage 3, number, ratio, fluency, reasoning and problem solving. Year 9 allocation is editorial within KS3. Numerical models, tasks and feedback are original; images retain actual Commons attribution and videos name their original publisher. Reviewed 12 November 2026.

PerpBisect.svg, Completerat, CC0. Unchanged original; caption and learning activities are original.

Perpendicular Bisector.gif, Chen-Pan Liao, CC0. Unchanged original; caption and learning activities are original.