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This lesson covers ruler-and-compass constructions, loci (the set of points satisfying a given condition), and three-figure bearings. These are practical geometry skills tested in Edexcel GCSE Mathematics, often requiring you to draw accurately with a compass and ruler.
| Term | Meaning |
|---|---|
| Construction | A geometric drawing made using only a compass and straight edge (ruler without measurements) |
| Locus (plural: loci) | The set of all points that satisfy a given condition |
| Bearing | A direction measured clockwise from north, given as a three-figure number (e.g. 045°, 210°) |
| Perpendicular bisector | A line at right angles to a line segment that passes through its midpoint |
| Angle bisector | A line that divides an angle exactly in half |
| Equidistant | The same distance away |
You must be able to perform these constructions using only a pair of compasses and a ruler (straight edge). Construction marks (arcs) must be shown — do not erase them.
This construction finds the midpoint and draws a line at right angles.
Method:
This line is the perpendicular bisector. Every point on it is equidistant from A and B.
This construction divides an angle exactly in half.
Method:
Method:
Method:
Method:
Follow the same method as constructing an equilateral triangle — the angle at A (or B) is 60°.
Construct the perpendicular bisector of a line segment — the resulting angle is 90°.
A locus is the set of all points that satisfy a particular condition. In GCSE questions, you typically shade a region or draw a line/curve.
| Condition | Locus |
|---|---|
| Fixed distance from a point | Circle (radius = given distance) |
| Fixed distance from a line | Two parallel lines either side, with semicircles at each end |
| Equidistant from two points | Perpendicular bisector of the line joining the two points |
| Equidistant from two lines that meet at a point | Angle bisector of the angle between the two lines |
A treasure is buried within 5 m of a tree T and closer to fence AB than fence CD (where AB and CD meet at a point). Shade the region where the treasure could be.
Step 1: Draw a circle of radius 5 m (to scale) centred on T. Step 2: Construct the angle bisector of the angle between AB and CD. Step 3: Shade the region that is inside the circle AND on the AB side of the bisector.
A goat is tied to a post P by a 6 m rope. Shade the region the goat can reach.
The goat can reach all points within 6 m of P — so the locus is a circle of radius 6 m centred on P.
Find the locus of points equidistant from A and B, where A and B are 8 cm apart.
Construct the perpendicular bisector of AB. Every point on this line is the same distance from A as from B.
Exam questions often combine two or more conditions. The answer is the region where all conditions are satisfied simultaneously (the intersection).
In a rectangular garden ABCD (12 m by 8 m), a tree must be:
Draw a line parallel to AB, 3 m away (inside the garden). Draw an arc of radius 5 m centred on C. The region satisfying both conditions is where these overlap.
A bearing is a direction measured clockwise from north. Bearings are always written as three figures.
| Direction | Bearing |
|---|---|
| North | 000° |
| East | 090° |
| South | 180° |
| West | 270° |
| North-East | 045° |
| South-West | 225° |
The bearing of B from A is 065°. Find the bearing of A from B.
The bearing of A from B = 065° + 180° = 245°
Rule: If the bearing from A to B is x°, the bearing from B to A is x° + 180° (if x < 180°) or x° - 180° (if x >= 180°).
A ship sails from P on a bearing of 130° for 50 km to Q, then on a bearing of 230° for 40 km to R. Find angle PQR.
At Q, the angle between the direction from P (reverse bearing = 130° + 180° = 310°) and the direction to R (230°) is: 310° - 230° = 80°
Town B is due east of town A. Town C is on a bearing of 140° from A. Find angle BAC.
Bearing of B from A = 090° (due east). Bearing of C from A = 140°. Angle BAC = 140° - 90° = 50°
Many bearing questions require a scale drawing. Use a protractor to measure angles from north and a ruler to measure distances to scale.
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