High School Maths · Year 2 · Unit 15 · Lesson 3 of 5
Loci: Find Every Point That Follows the Rule
A goat is tied to a post T in the middle of a large field with a rope 6 m long.
Which parts of the field can the goat reach?
Every point up to 6 m from the post: a circle of radius 6 m and everything inside it. The points exactly 6 m from T form the circle itself. A set of points that obeys a rule like this is called a locus (plural: loci).
Master the essentials
Locus
The set of ALL the points that obey a rule. It can be a line, a curve or a region.
Plural: loci
A fixed distance from a point
A circle, with the point as its centre.
Points 3 cm from P: a circle of radius 3 cm
Within a distance
"Less than" a distance from a point is the inside of the circle. Draw the circle dashed when its edge is not included.
Less than 3 cm from P: inside the circle
A fixed distance from a line segment
Two lines parallel to the segment, joined by a semicircle round each end.
It looks like a running track
A fixed distance from an endless line
Two parallel lines, one on each side. There are no ends, so no semicircles.
Like the two edges of a road
The same distance from two points
The perpendicular bisector of the line joining the points.
Every point on it is equally far from A and B
The same distance from two lines
The bisector of the angle between the lines. Two crossing lines have two bisectors.
Construct it with compasses
Nearer to A than to B
All the points on A's side of the perpendicular bisector of AB.
The bisector itself is the boundary
See it done
1. Describe the locus of points exactly 2.5 cm from a line segment AB that is 6 cm long, and find its total length. Leave your answer in terms of π.
- Beside AB: two straight lines parallel to AB, 2.5 cm away, each 6 cm long.
- Round each end: a semicircle of radius 2.5 cm, centred on A or on B.
- Straight parts: 2 × 6 = 12 cm. The two semicircles make one circle: 2 × π × 2.5 = 5π cm.
- Total length: (12 + 5π) cm.
(12 + 5π) cm
2. Two straight walls meet at a corner O at an angle of 70°. A sprinkler must be the same distance from both walls. Describe where it can go.
- Points the same distance from two lines lie on the bisector of the angle between them.
- Construct the bisector of the 70° angle with compasses.
- The bisector makes 70° ÷ 2 = 35° with each wall.
Anywhere on the bisector of the angle, at 35° to each wall
3. P and Q are 8 cm apart. Find the points that are 5 cm from P and also 5 cm from Q, and the distance between them.
- 5 cm from P: a circle, centre P, radius 5 cm. The same for Q.
- The circles cross at two points, X and Y. Both are on the perpendicular bisector of PQ, which meets PQ at its midpoint M, 4 cm from P.
- XM² = 5² − 4² = 25 − 16 = 9, so XM = 3 cm.
- X and Y are on opposite sides of PQ: XY = 3 + 3 = 6 cm.
Two points, 6 cm apart
Common mistake: Forgetting the curved ends. Points a fixed distance from a line SEGMENT make two straight lines AND two semicircles round the ends. Only an endless line gives just two parallel lines. And read the rule carefully: "exactly" gives a line or curve, but "less than" or "at most" gives a region.
10 questions with instant feedback
Choose the correct answer. For each rule, picture the locus first: a circle, a pair of lines, a running track or a bisector? Questions 1–3 are warm-ups, 4–7 are at exam standard and 8–10 are challenge questions. After you answer, the feedback shows the full solution.
Warm-up (The basic skill: one or two steps.)
1.Describe the locus of points that are exactly 7 cm from a fixed point O.
Not quite. That is the locus of points UP TO 7 cm from O. "Exactly 7 cm" is just the circle.
Solution
- Every point exactly 7 cm from O lies on a circle round O.
- The radius of the circle is 7 cm.
Answer: B. A circle with centre O and radius 7 cm.
Correct.
Solution
- Every point exactly 7 cm from O lies on a circle round O.
- The radius of the circle is 7 cm.
Not quite. Its points are only 3.5 cm from O.
Solution
- Every point exactly 7 cm from O lies on a circle round O.
- The radius of the circle is 7 cm.
Answer: B. A circle with centre O and radius 7 cm.
Not quite. Parallel lines are the locus for a fixed distance from a LINE, not from a point.
Solution
- Every point exactly 7 cm from O lies on a circle round O.
- The radius of the circle is 7 cm.
Answer: B. A circle with centre O and radius 7 cm.
2.Points A and B are 9 cm apart. Which locus is the set of points that are the same distance from A as from B?
Not quite. From points on this circle, AB makes a right angle. They are not equally far from A and B.
Solution
- Equal arcs from A and from B cross at points that are equally far from A and B.
- All such points lie on one straight line: the perpendicular bisector of AB.
Answer: D. The perpendicular bisector of AB.
Not quite. Only one point of this line, the midpoint of AB, is equally far from A and B.
Solution
- Equal arcs from A and from B cross at points that are equally far from A and B.
- All such points lie on one straight line: the perpendicular bisector of AB.
Answer: D. The perpendicular bisector of AB.
Not quite. These points are 9 cm from the midpoint, not equally far from A and B.
Solution
- Equal arcs from A and from B cross at points that are equally far from A and B.
- All such points lie on one straight line: the perpendicular bisector of AB.
Answer: D. The perpendicular bisector of AB.
Correct.
Solution
- Equal arcs from A and from B cross at points that are equally far from A and B.
- All such points lie on one straight line: the perpendicular bisector of AB.
3.A locus is drawn around a line segment PQ. It is made of two lines parallel to PQ and the same length as PQ, joined by a semicircle round P and a semicircle round Q. Which rule gives this locus?
Not quite. That locus is a circle. This shape has straight sides.
Solution
- Beside PQ, the locus is two straight lines parallel to PQ.
- Beyond P and beyond Q, it curves round in semicircles.
- That is the locus of points a fixed distance from the segment PQ.
Answer: D. Points that are a fixed distance from the line segment PQ.
Not quite. That locus is just two parallel lines, with no curved ends.
Solution
- Beside PQ, the locus is two straight lines parallel to PQ.
- Beyond P and beyond Q, it curves round in semicircles.
- That is the locus of points a fixed distance from the segment PQ.
Answer: D. Points that are a fixed distance from the line segment PQ.
Not quite. That locus is a single straight line: the perpendicular bisector of PQ.
Solution
- Beside PQ, the locus is two straight lines parallel to PQ.
- Beyond P and beyond Q, it curves round in semicircles.
- That is the locus of points a fixed distance from the segment PQ.
Answer: D. Points that are a fixed distance from the line segment PQ.
Correct.
Solution
- Beside PQ, the locus is two straight lines parallel to PQ.
- Beyond P and beyond Q, it curves round in semicircles.
- That is the locus of points a fixed distance from the segment PQ.
Exam standard (Several steps, like real exam questions.)
4.Lines OA and OB meet at O, and angle AOB = 60°. P is inside the angle, the same distance from OA as from OB, and OP = 14 cm. How far is P from OA?
Not quite. This is OP itself, not the distance from P to the line OA.
Solution
- P is the same distance from both lines, so P is on the bisector: angle AOP = 30°.
- Draw PN at 90° to OA. Triangle OPN has angles of 30°, 60° and 90°.
- That is half of an equilateral triangle of side OP = 14 cm, so PN = 14 ÷ 2 = 7 cm.
Answer: B. 7 cm.
Correct.
Solution
- P is the same distance from both lines, so P is on the bisector: angle AOP = 30°.
- Draw PN at 90° to OA. Triangle OPN has angles of 30°, 60° and 90°.
- That is half of an equilateral triangle of side OP = 14 cm, so PN = 14 ÷ 2 = 7 cm.
Not quite. This is ON, the distance along OA to the foot of the perpendicular.
Solution
- P is the same distance from both lines, so P is on the bisector: angle AOP = 30°.
- Draw PN at 90° to OA. Triangle OPN has angles of 30°, 60° and 90°.
- That is half of an equilateral triangle of side OP = 14 cm, so PN = 14 ÷ 2 = 7 cm.
Answer: B. 7 cm.
Not quite. This doubles OP. The distance to OA is a SHORTER side of the right-angled triangle.
Solution
- P is the same distance from both lines, so P is on the bisector: angle AOP = 30°.
- Draw PN at 90° to OA. Triangle OPN has angles of 30°, 60° and 90°.
- That is half of an equilateral triangle of side OP = 14 cm, so PN = 14 ÷ 2 = 7 cm.
Answer: B. 7 cm.
5.A and B are 16 cm apart. Two points are each 10 cm from A and 10 cm from B. How far apart are these two points?
Not quite. This is the distance from one of the points to AB. The other point is on the other side.
Solution
- Both points are on the perpendicular bisector of AB, which meets AB at its midpoint M, 8 cm from A.
- For each point X: XM² = 10² − 8² = 100 − 64 = 36, so XM = 6 cm.
- The points are on opposite sides of AB: 6 + 6 = 12 cm apart.
Answer: C. 12 cm.
Not quite. This adds the squares. The 10 cm side is the hypotenuse, so subtract.
Solution
- Both points are on the perpendicular bisector of AB, which meets AB at its midpoint M, 8 cm from A.
- For each point X: XM² = 10² − 8² = 100 − 64 = 36, so XM = 6 cm.
- The points are on opposite sides of AB: 6 + 6 = 12 cm apart.
Answer: C. 12 cm.
Correct.
Solution
- Both points are on the perpendicular bisector of AB, which meets AB at its midpoint M, 8 cm from A.
- For each point X: XM² = 10² − 8² = 100 − 64 = 36, so XM = 6 cm.
- The points are on opposite sides of AB: 6 + 6 = 12 cm apart.
Not quite. This subtracts the lengths: 2 × (10 − 8). Use Pythagoras.
Solution
- Both points are on the perpendicular bisector of AB, which meets AB at its midpoint M, 8 cm from A.
- For each point X: XM² = 10² − 8² = 100 − 64 = 36, so XM = 6 cm.
- The points are on opposite sides of AB: 6 + 6 = 12 cm apart.
Answer: C. 12 cm.
6.The table shows how far four points are from a point A and from a point B. Which point lies on the perpendicular bisector of AB?
| Point | Distance from A (cm) | Distance from B (cm) |
|---|---|---|
| P | 5.2 | 5.8 |
| Q | 6.4 | 6.4 |
| R | 7 | 3.5 |
| S | 4.9 | 9.4 |
Not quite. P is 5.2 cm from A but 5.8 cm from B, so it is nearer to A.
Solution
- A point on the perpendicular bisector of AB is the same distance from A as from B.
- Only Q has equal distances: 6.4 cm and 6.4 cm.
Answer: C. Q.
Not quite. R is twice as far from A as from B.
Solution
- A point on the perpendicular bisector of AB is the same distance from A as from B.
- Only Q has equal distances: 6.4 cm and 6.4 cm.
Answer: C. Q.
Correct.
Solution
- A point on the perpendicular bisector of AB is the same distance from A as from B.
- Only Q has equal distances: 6.4 cm and 6.4 cm.
Not quite. S has the digits swapped, but 4.9 cm and 9.4 cm are not equal.
Solution
- A point on the perpendicular bisector of AB is the same distance from A as from B.
- Only Q has equal distances: 6.4 cm and 6.4 cm.
Answer: C. Q.
7.Find the area of the region of points that are within 3 cm of a line segment 8 cm long. Leave your answer in terms of π.
Not quite. This counts one side of the segment only. The region is 3 cm wide on EACH side.
Solution
- The region is a rectangle 8 cm by 6 cm (3 cm on each side), with a semicircle of radius 3 cm at each end.
- Rectangle: 8 × 6 = 48 cm².
- Two semicircles make one circle: π × 3² = 9π cm². Total: (48 + 9π) cm².
Answer: D. (48 + 9π) cm².
Not quite. The two ends make two semicircles, which together make one circle: 9π, not 18π.
Solution
- The region is a rectangle 8 cm by 6 cm (3 cm on each side), with a semicircle of radius 3 cm at each end.
- Rectangle: 8 × 6 = 48 cm².
- Two semicircles make one circle: π × 3² = 9π cm². Total: (48 + 9π) cm².
Answer: D. (48 + 9π) cm².
Not quite. This uses 2πr, which is a length. Area needs πr².
Solution
- The region is a rectangle 8 cm by 6 cm (3 cm on each side), with a semicircle of radius 3 cm at each end.
- Rectangle: 8 × 6 = 48 cm².
- Two semicircles make one circle: π × 3² = 9π cm². Total: (48 + 9π) cm².
Answer: D. (48 + 9π) cm².
Correct.
Solution
- The region is a rectangle 8 cm by 6 cm (3 cm on each side), with a semicircle of radius 3 cm at each end.
- Rectangle: 8 × 6 = 48 cm².
- Two semicircles make one circle: π × 3² = 9π cm². Total: (48 + 9π) cm².
Challenge (Harder problems that mix ideas or need reasoning.)
8.Two straight lines cross, making angles of 50° and 130°. Which working shows the angle between the two lines that make up the locus of points equidistant from both lines?
Correct.
Solution
- Points equidistant from two crossing lines lie on the bisectors of the angles between them.
- Between one bisector and a line: 50° ÷ 2 = 25°. Between that line and the other bisector: 130° ÷ 2 = 65°.
- 25° + 65° = 90°. The two bisectors are always at right angles.
Not quite. 25° is the angle between one bisector and one of the LINES, not between the two bisectors.
Solution
- Points equidistant from two crossing lines lie on the bisectors of the angles between them.
- Between one bisector and a line: 50° ÷ 2 = 25°. Between that line and the other bisector: 130° ÷ 2 = 65°.
- 25° + 65° = 90°. The two bisectors are always at right angles.
Answer: A. The bisectors split 50° into 25° + 25° and 130° into 65° + 65°. So the angle between them is 25° + 65° = 90°.
Not quite. 65° is the angle between the other bisector and a line. Add the 25° on the other side of that line.
Solution
- Points equidistant from two crossing lines lie on the bisectors of the angles between them.
- Between one bisector and a line: 50° ÷ 2 = 25°. Between that line and the other bisector: 130° ÷ 2 = 65°.
- 25° + 65° = 90°. The two bisectors are always at right angles.
Answer: A. The bisectors split 50° into 25° + 25° and 130° into 65° + 65°. So the angle between them is 25° + 65° = 90°.
Not quite. Each bisector takes HALF of its angle, so they meet at 25° + 65°, not 180°.
Solution
- Points equidistant from two crossing lines lie on the bisectors of the angles between them.
- Between one bisector and a line: 50° ÷ 2 = 25°. Between that line and the other bisector: 130° ÷ 2 = 65°.
- 25° + 65° = 90°. The two bisectors are always at right angles.
Answer: A. The bisectors split 50° into 25° + 25° and 130° into 65° + 65°. So the angle between them is 25° + 65° = 90°.
9.The locus of points d cm from a line segment 12 cm long has a total length of (24 + 14π) cm. Find d.
Correct.
Solution
- The locus is two straight parts, each 12 cm, and a semicircle of radius d at each end.
- Straight parts: 2 × 12 = 24 cm. Two semicircles make one circle: 2πd.
- 2πd = 14π, so d = 7.
Not quite. This is 2d. The two semicircles make one circle of length 2πd = 14π.
Solution
- The locus is two straight parts, each 12 cm, and a semicircle of radius d at each end.
- Straight parts: 2 × 12 = 24 cm. Two semicircles make one circle: 2πd.
- 2πd = 14π, so d = 7.
Answer: A. 7 cm.
Not quite. This is the length of the segment.
Solution
- The locus is two straight parts, each 12 cm, and a semicircle of radius d at each end.
- Straight parts: 2 × 12 = 24 cm. Two semicircles make one circle: 2πd.
- 2πd = 14π, so d = 7.
Answer: A. 7 cm.
Not quite. This uses πd², which is an area. The length of a circle is 2πd.
Solution
- The locus is two straight parts, each 12 cm, and a semicircle of radius d at each end.
- Straight parts: 2 × 12 = 24 cm. Two semicircles make one circle: 2πd.
- 2πd = 14π, so d = 7.
Answer: A. 7 cm.
10.A point moves outside a square of side 6 cm, always exactly 3 cm from the nearest point of the square. Find the length of its path. Leave your answer in terms of π.
Not quite. This draws a bigger square with sharp corners. Round each corner the point moves on a quarter circle.
Solution
- Along each side the path is straight and 6 cm long: 4 × 6 = 24 cm.
- Round each corner it moves on a quarter circle of radius 3 cm.
- Four quarter circles make one circle: 2 × π × 3 = 6π cm. Total: (24 + 6π) cm.
Answer: C. (24 + 6π) cm.
Not quite. Four quarter circles make ONE whole circle: 2 × π × 3 = 6π.
Solution
- Along each side the path is straight and 6 cm long: 4 × 6 = 24 cm.
- Round each corner it moves on a quarter circle of radius 3 cm.
- Four quarter circles make one circle: 2 × π × 3 = 6π cm. Total: (24 + 6π) cm.
Answer: C. (24 + 6π) cm.
Correct.
Solution
- Along each side the path is straight and 6 cm long: 4 × 6 = 24 cm.
- Round each corner it moves on a quarter circle of radius 3 cm.
- Four quarter circles make one circle: 2 × π × 3 = 6π cm. Total: (24 + 6π) cm.
Not quite. This counts a whole circle at every corner. Each corner has only a quarter circle.
Solution
- Along each side the path is straight and 6 cm long: 4 × 6 = 24 cm.
- Round each corner it moves on a quarter circle of radius 3 cm.
- Four quarter circles make one circle: 2 × π × 3 = 6π cm. Total: (24 + 6π) cm.
Answer: C. (24 + 6π) cm.
Where is the treasure?
A treasure map shows a tree T and a well W, 6 m apart. Answer each question on paper, with a sketch, then compare with the model.
1. The treasure is exactly 4 m from the tree T. Describe this locus.
2. The treasure is also the same distance from T as from W. Describe this locus, and say how many points obey both rules.
3. Find the distance between those points. Leave your answer as a surd.
- A circle with centre T and radius 4 m.
- The perpendicular bisector of TW.
- It passes 3 m from T (half of TW), and 3 m is less than 4 m, so it cuts the circle at two points.
- Each point X is on the bisector, which meets TW at its midpoint M, with TM = 3 m. XM² = 4² − 3² = 16 − 9 = 7, so XM = √7 m.
- The two points are on opposite sides of TW: the distance between them is 2√7 m.
Next: the Lesson 3 exit quiz. Five questions, recorded on your dashboard. Aim for 4 out of 5.