OCR GCSE Mathematics: Ruler-and-compass constructions and loci (J560)

This study guide follows section 8.02 (Ruler and compass constructions) of the OCR GCSE (9-1) Mathematics J560 specification, Version 2.1 (August 2026): 8.02a to 8.02d. All of it is Foundation and Higher content, so no part of it is restricted to the Higher tier. The specification says any content may be assessed on either the calculator or the non-calculator papers. Constructions are done by hand, with a ruler and a pair of compasses.

Use it with the revision notes and the practice questions. The OCR GCSE Mathematics hub, the printable checklist and the 10-minute diagnostics help you plan. The exam preparation page covers general paper technique.

What this unit covers

Ref	What you must be able to do	Tier
8.02a	Construct the perpendicular bisector and midpoint of a line segment	Foundation and Higher
8.02b	Construct the bisector of an angle formed from two lines	Foundation and Higher
8.02c	Construct the perpendicular from a point to a line, and the perpendicular to a line at a point; know that the perpendicular distance is the shortest distance	Foundation and Higher
8.02d	Apply constructions to draw figures and identify loci, including real-world problems; understand the term “equidistant”	Foundation and Higher

Equipment and habits

You need a sharp pencil, a ruler, and compasses that do not slip. The specification also expects you to use a protractor to construct and measure angles and a ruler to draw and measure lines, which you will need to set up the starting diagram.

• Draw the starting lines lightly, then draw the final answer firmly.
• Keep every construction arc. Arcs show how you found the line.
• Do not change the compass setting in the middle of one step.
• Make arcs long enough to cross clearly.

A line segment has two end points. A locus (plural loci) is the set of all points that follow a rule. Equidistant means the same distance away.

8.02a: Perpendicular bisector and midpoint

The perpendicular bisector of a segment AB cuts AB in half and meets it at 90°. The point where it crosses AB is the midpoint.

Method

1. Open the compasses to more than half the length of AB. Keep this setting.
2. Put the point on A and draw an arc above and below AB.
3. Put the point on B and draw two arcs with the same radius, crossing the first two.
4. Join the two crossing points with a ruler. This line is the perpendicular bisector.
5. Label where it crosses AB as M, the midpoint.

        X
        |
        |
A ------M------ B
        |
        |
        Y

Why it works. Both arcs have the same radius, so AX = BX = AY = BY. Every point on the line XY is the same distance from A as from B. It is the locus of points equidistant from A and B.

Worked example 1. Construct the midpoint of a segment AB of length 7.6 cm.

1. Draw AB = 7.6 cm.
2. Set the compasses to 5 cm, which is more than half of 7.6 cm.
3. Draw arcs from A, above and below AB. Repeat from B with the same 5 cm radius.
4. Join the crossing points. Where this line crosses AB is M.
5. Check: AM and MB should each measure 3.8 cm. If they differ by more than about 1 mm, redraw.

8.02b: Angle bisector

The angle bisector splits an angle into two equal angles. Here the angle is formed by two lines meeting at a point V.

Method

1. Put the compass point on V and draw an arc that crosses both lines, at P and Q.
2. Put the point on P and draw an arc inside the angle.
3. Keep the same radius, put the point on Q and draw a second arc crossing the first at R.
4. Join V to R with a ruler. Extend it if needed.

        R
       /
  P   /
   \ /
    V ------ Q   (arcs from P and Q cross at R)

Why it works. VP = VQ and PR = QR, and VR is shared. The two triangles VPR and VQR match, so the two angles at V are equal.

Every point on the bisector is equidistant from the two lines. This gives the locus of points the same distance from two intersecting lines.

Worked example 2. Bisect an angle of 84°.

1. Draw an 84° angle with a protractor.
2. Draw an arc from V, crossing both lines.
3. Draw crossing arcs from the two crossing points.
4. Join V to the crossing point.
5. Check with the protractor: each part should be 42°.

8.02c: Perpendiculars and the shortest distance

You must be able to do two different constructions.

Perpendicular from a point to a line

The point P is not on the line. You are dropping a perpendicular from P.

1. Put the compass point on P and draw an arc that cuts the line twice, at A and B.
2. Put the point on A and draw an arc on the far side of the line from P.
3. Keep the same radius, put the point on B and draw a second arc crossing the first at X.
4. Join P to X. The line PX meets the original line at 90°.

Perpendicular at a point on a line

The point T is on the line.

1. Put the compass point on T and draw arcs cutting the line either side, at A and B (equal distances from T).
2. Open the compasses wider. Draw arcs from A and B that cross, above the line, at X.
3. Join T to X. The line TX is perpendicular to the line at T.

This is the perpendicular bisector of AB again, because T is the midpoint of AB.

The shortest distance

The perpendicular distance from a point to a line is the shortest distance from the point to the line. Any other route is the hypotenuse of a right-angled triangle, and the hypotenuse is the longest side.

Worked example 3. The foot of the perpendicular from P to a line is F, and PF = 3.6 cm. A point X is on the line with FX = 4.8 cm. Find PX and compare.

PX² = PF² + FX²
    = 3.6² + 4.8²
    = 12.96 + 23.04
    = 36
PX  = 6 cm

PX = 6 cm is longer than PF = 3.6 cm. So the shortest distance from P to the line is 3.6 cm.

Worked example 4. Construct a right angle at the point T on a horizontal line, then use it to draw a vertical segment TX of length 5 cm.

1. Draw arcs from T cutting the line at A and B, each 2 cm from T.
2. Draw arcs of radius 3 cm from A and B. They cross above T at Y.
3. Draw the line through T and Y.
4. Measure 5 cm from T along it and mark X.

8.02d: Loci

A locus is a rule turned into a drawing. First decide which rule applies, then draw the line or region.

Rule	Locus
A fixed distance from a point	A circle, centre the point
Within a fixed distance of a point	The inside of a circle
Equidistant from two points A and B	The perpendicular bisector of AB
Equidistant from two intersecting lines	The angle bisector
A fixed distance from a line segment	A “racetrack” shape: two parallel lines with a semicircle at each end
Equidistant from two parallel lines	A parallel line half-way between them

Locus around a segment. All the points exactly 2 cm from a segment CD form two straight lines 2 cm away and parallel to CD, joined by a semicircle of radius 2 cm at each end.

     ____________
    (            )     <- the outline is 2 cm from CD
    ( C -------- D )
    (____________)

Strict and non-strict boundaries

• “Less than 3 cm” or “closer than”: draw the boundary with a dashed line and do not include it.
• “At most 3 cm” or “no further than”: draw the boundary with a solid line and include it.

Shade the region the question asks for. Say clearly in the key or a label what the shading means.

Worked example 5. A rectangle PQRS has PQ = 9 m and PS = 6 m. Scale: 1 cm represents 1 m. A tree must be placed (i) closer to P than to Q, (ii) closer to side PQ than to side PS, (iii) no more than 6 m from P. Show the region.

Take P as the origin with Q to the right and S above.

1. For (i), construct the perpendicular bisector of PQ. It is the line 4.5 cm from P. The tree is to the left of it.
2. For (ii), construct the bisector of angle P, which is 90°, so the bisector makes 45° with PQ. Points closer to PQ than PS are below this line.
3. For (iii), draw an arc of radius 6 cm centred on P. The tree is inside it or on it.
4. The region has these corners: P, the point where the bisector meets the arc, the point where the arc meets the perpendicular bisector, and the foot of the perpendicular bisector on PQ.

Checking the numbers: the bisector meets the arc at 6 ÷ √2 = 4.24 cm along and up. The arc meets the perpendicular bisector 3.97 cm above PQ, because √(6² − 4.5²) = √15.75 = 3.97. The region therefore has four corners, one edge of which is the curved arc, and its right-hand edge is 3.97 cm high.

Worked example 6. A signal mast M is at one end of a straight road AB. Mark all places within 3 km of M that are also more than 2 km from the road. Scale: 1 cm for 1 km, AB = 8 km, M at A.

1. Draw the circle of radius 3 cm centred on A.
2. Draw two dashed lines 2 cm either side of AB (parallel to it) and a dashed semicircle of radius 2 cm round A, because “more than 2 km” is a strict condition.
3. The road is a segment, so the points within 2 km of it also include a semicircle of radius 2 cm round A. The required places are inside the 3 cm circle and outside this racetrack shape, which leaves one curved band round the left end of the road.

Real-world problems

Most loci problems are set on a scale drawing. Convert each distance to centimetres first, draw the plan, then read off the answer and convert back.

Worked example 7. A garden is 12 m by 8 m. A lamp at the corner A lights everything within 5 m. A pond is at the midpoint of the opposite long side. Use scale 1 cm to 2 m. Is the pond lit?

1. Scale: 12 m is 6 cm, 8 m is 4 cm. The lamp radius is 5 m = 2.5 cm.
2. Let A be at the bottom left. The pond is at the midpoint of the top long side, 3 cm along and 4 cm up from A. Its distance from A is 5 cm on the drawing, which is √(3² + 4²) = 5.
3. The drawing distance 5 cm represents 10 m. This is more than 5 m, so the pond is not lit.

Common errors

• Changing the compass setting between the two arcs in step 3 of a bisector.
• Using a radius less than half of AB, so the arcs never meet.
• Drawing a perpendicular bisector freehand through the “middle” and not using arcs.
• Measuring shortest distance along a slanted line rather than the perpendicular.
• Forgetting that the locus around a segment has rounded ends.
• Shading the wrong side of a boundary.
• Not converting the scale before drawing.

Next steps

Try the practice questions, then use the revision notes to check recall. Return to the OCR GCSE Mathematics hub for other topics.

Official syllabus

OCR GCSE (9-1) Mathematics J560 specification, Version 2.1 (August 2026), section 8.02, published by OCR.