OxfordAQA International GCSE Mathematics: Scale drawings, bearings, constructions and loci (9260) – Revision Notes

These notes condense scale drawings, bearings, constructions and loci for the OxfordAQA International GCSE Mathematics (9260) specification, Version 5.1 (for exams May/June 2018 onwards), references G12 and G13 in section 3.3.1. Both references are Core content, which is also assessable on the Extension papers, and this unit has no Extension-only statements. For full explanations see the study guide. To test yourself, go to the practice set, the course hub, the printable checklist and the free 10-minute diagnostics.

Key facts at a glance

Idea	Rule
Bearing	Angle clockwise from north, three figures (007°, 045°, 310°)
Compass points	N 000°, NE 045°, E 090°, SE 135°, S 180°, SW 225°, W 270°, NW 315°
Back bearing	Add 180° if below 180°, otherwise subtract 180°
Scale ratio	1 : n means 1 unit on the drawing is n units in real life
Drawing to real	Multiply by n, then convert units
Real to drawing	Divide by n, after converting to the same unit
Locus	Every point that obeys a given rule
Shortest route, point to line	Along the perpendicular

Measuring (G12)

• Start a ruler reading from zero and a protractor reading from the 0° mark on the arm you line up.
• Check the reading against the picture: acute means below 90°, obtuse means between 90° and 180°.
• Reflex angle = 360° − the angle you measured. For example, a measured 128° gives a reflex angle of 232°.
• Drawings are never perfect, so work to 1 mm and 1° and show your measured values.

Scales and maps (G12)

Method in steps:

1. Express the scale as 1 : n with both sides in one unit (convert to cm).
2. Multiply to go from the drawing to real life, divide to go the other way.
3. Convert the answer to a sensible unit (100 cm = 1 m, 1000 m = 1 km, 100 000 cm = 1 km).

Worked reminders:

Scale 1 : 1500, drawing 9 cm      9 × 1500 = 13 500 cm = 135 m
Scale 1 : 90 000, real 3.6 km     360 000 cm ÷ 90 000 = 4 cm
"1 cm represents 400 m"           400 m = 40 000 cm, so 1 : 40 000
Scale 1 : 600 000, drawing 7.5 cm 7.5 × 600 000 = 4 500 000 cm = 45 km

If a question gives a statement such as “3 cm represents 6 km”, scale it down to “1 cm represents 2 km” and then convert 2 km to cm.

Bearings (G12)

Method in steps:

1. Mark the point you are standing at. The word after “from” names this point.
2. Draw a north line there.
3. Measure clockwise from the north line to the line that joins the two points.
4. Write three figures.

For a journey with several legs, repeat from step 2 at every turning point.

Bearing of B from A = 126°   →   bearing of A from B = 126° + 180° = 306°
Bearing of B from A = 287°   →   bearing of A from B = 287° − 180° = 107°

Why the back bearing works: north lines are parallel, so the angle between the two directions is a half turn. If a bearing is above 180°, say 342°, subtracting 180° gives 162°.

Angles in a diagram at a turning point often come from parallel lines. The angles between a north line and a path on the same side add to 180°. That is useful when you are told to calculate rather than draw.

Constructions (G13)

Every construction needs the two arcs to have the same compass width and you must leave the arcs visible.

Construction	Arcs and lines
Perpendicular bisector of AB	Equal-width arcs from A and from B (wider than half of AB), above and below; join the crossings
Perpendicular dropped from point P	Arc centred P cuts the line twice; arcs from those two cuts cross; join the crossing to P
Perpendicular erected at P on the line	Arcs centred P cut the line either side; bisect the gap between the cuts
Bisector of angle at V	Arc from V cuts both arms; arcs from the two cut points cross; join to V
60° at P	Arc from P cuts the line at X; arc from X with the same width cuts the first arc at Z; join P to Z

Other angles come from these. 30° = half of 60°. 120° = two 60° angles. 90° = perpendicular. 45° = half of 90°.

To construct a triangle from three sides, draw one side, then draw arcs from each end with the other two lengths. The arcs cross at the third vertex.

Loci (G13)

Condition	Shape
Fixed distance r from point O	Circle centre O, radius r
Fixed distance r from segment AB	Two lines parallel to AB, distance r away, joined by a semicircle at each end
Same distance from A as from B	Perpendicular bisector of AB
Same distance from two crossing lines	Bisector of the angle between them
Closer to A than B	A’s side of the perpendicular bisector
Within a distance r of a point	Inside the circle

Method in steps for a region:

1. Draw each boundary lightly with pencil.
2. For each, decide which side is wanted.
3. Test one point if you are unsure.
4. Shade only the part that meets every rule.

Use a dashed line for a boundary that is not included (a strict inequality, such as “more than 4 m from”) and a solid line when points on it count, but check what your question asks.

Worked reminder: a combined locus

A path PQ is 10 m long. A region is described as “less than 3 m from the path and nearer to P than to Q”. Draw a pill shape around PQ, 3 m out on each side with a semicircle at each end. Then draw the perpendicular bisector of PQ, 5 m from P, and shade only the half of the pill that holds P. A point 4 m along the path from P and 2 m to the side is in the region: it is on the P side of the bisector and 2 m is less than 3 m from the path.

Shortest route to a line (G13)

The shortest distance is along the perpendicular. To find it, construct the perpendicular from the point to the line and measure it. Example: a ship 8 km from a port on a bearing of 040°, with a coast running west to east through the port, is 8 × cos 40° = 6.13 km from the coast.

Exam technique

• Read the question for the word “construct”. It means ruler and compasses, with arcs left on the page.
• For a scale drawing, write the scale you chose at the top of the working.
• State units in every answer: km, m, cm or degrees.
• Where a question asks you to calculate, use the bearings to find the angles in a triangle and then use the right-angled or non-right-angled triangle methods from other topics.
• Check that a bearing answer lies between 000° and 360°.

Must-know distinctions

• “Bearing of B from A” is not the same as “bearing of A from B”. They differ by 180°.
• A perpendicular bisector cuts a line in half at 90°. An angle bisector cuts an angle in half. The arcs look similar, the centres differ.
• Equidistant from two points gives a line. Fixed distance from a point gives a circle.
• Scale ratio does not carry a unit, but you must use the same unit on both sides when you write it.

Quick self-test

1. Write the three-figure bearing for south-east.
2. The bearing of N from M is 038°. Find the bearing of M from N.
3. A scale is 1 : 40 000. A path is 6 cm long on the map. How long is it in real life?
4. What does the locus “equidistant from two points” look like?
5. How would you construct an angle of 30° with ruler and compasses?
6. A drawing uses the scale 1 : 200 000. What drawing length shows 10 km?
7. The smaller angle at a vertex measures 128° on a protractor. Find the reflex angle there.
8. The bearing of B from A is 287°. What is the bearing of A from B?
9. Point P is 7 cm from a line, measured along the perpendicular. Another point R on the line is 9 cm from P. Which length is the shortest distance from P to the line, and why?
10. Describe the locus of points less than 2 cm from a point O.

Answers

1. 135°
2. 038° + 180° = 218°
3. 6 × 40 000 = 240 000 cm = 2.4 km
4. A straight line, the perpendicular bisector of the segment joining the points.
5. Construct 60°, then bisect it.
6. 10 km = 1 000 000 cm; 1 000 000 ÷ 200 000 = 5 cm
7. 360° − 128° = 232°
8. 287° − 180° = 107°
9. 7 cm. The perpendicular is the shortest route to the line, and any other route, such as PR, is longer.
10. The inside of a circle, centre O, radius 2 cm (the circle itself not included).

Where marks are usually lost

• Writing a bearing with two figures or with the unit missing.
• Putting the north line at the wrong end of the journey.
• Mixing km and cm in a scale calculation.
• Using the wrong protractor scale and reading 130° instead of 50°.
• Compass width changing between two arcs, so the crossing points are off.
• Leaving out the arcs, so there is no evidence you constructed rather than guessed.
• Shading a region that satisfies only some of the rules.
• Giving a slanted distance instead of the perpendicular as the shortest distance.

Official syllabus

These notes follow the OxfordAQA International GCSE Mathematics (9260) specification, Version 5.1, for exams May/June 2018 onwards, published by OxfordAQA, references G12 and G13.