Every point that obeys one rule
A locus is the set of all the points that obey one rule, drawn as a single shape. The plural is loci.
The simplest rule is "exactly r from a point C". Every point at that distance lies on the circle with center C and radius r, because that is what a circle is. Every point inside the circle is nearer than r, and every point outside is farther, so the circle holds exactly the points that obey the rule and no others. The locus of points exactly 2 units from C is the circle of radius 2 around C. It is not a square: a square’s corners are farther from its center than the middles of its sides.
Every point exactly r from the center lies on this circle, and every point on it is exactly r from the center.
Equally far from two points
The locus of points equally far from two points A and B is the perpendicular bisector of AB: the line through the midpoint of AB at a right angle to it. Every point on it is as far from A as from B. Every point off it is nearer to one of them.
It is drawn exactly with the construction: equal arcs from A and from B, and a line through the two crossings.
The perpendicular bisector of AB: every point on it is the same distance from A and from B.
A fixed distance from a line
The distance from a point to a line is measured along the perpendicular, the shortest way. The points exactly 3 m from a long straight line therefore make two lines parallel to it, 3 m away, one on each side.
A path kept exactly 3 m from a straight wall, on one side only, is one of those lines: a straight line parallel to the wall. It is not a circle: a circle is the locus for a fixed distance from one point, and here the distance is measured to a whole line.
The wall runs along the line y = 0. The path 3 m from it, on one side, is the line y = 3, parallel to the wall.
Equally far from the two arms of an angle
The locus of points equally far from the two arms of an angle is the angle bisector. The distance to each arm is measured along the perpendicular to that arm.
Take a point P on the bisector and drop the perpendicular from P to each arm. Fold the page along the bisector: the two arms land on each other, and P stays where it is. The perpendicular from P to one arm then lies along the perpendicular from P to the other arm, because there is only one line from P that meets an arm at a right angle. So the two distances are equal.
The bisector of angle V: every point on it is the same distance from the two arms.
Regions, and two rules at once
Some rules ask for more than a line. "Within 2 m of C" is every point on or inside the circle of radius 2 around C: a whole disc, not only its edge. "Nearer to A than to B" is every point on A’s side of the perpendicular bisector of AB. To check which side is which, test one point you know, such as A itself.
When a point must obey two rules at once, draw the locus for each rule. The points that obey both are where the two loci meet, or where the two regions overlap, and that part is shaded as the answer.
So a locus question ends with a drawing. A locus is the answer set, drawn, and the constructions are how it is drawn exactly: a circle with the compass, and the two bisectors with pairs of equal arcs.
A is at (−2, −2) and B is at (2, 2). Their perpendicular bisector is the line y = −x, and the shaded side, below it, holds every point nearer to A than to B. The line is dashed because its own points are equally far from both, so they are not nearer to A.
Near a segment
The points within 4 m of a straight segment take two kinds of shape. Beside the segment, the nearest point of it is straight across, along the perpendicular, so the points within 4 m make a strip 4 m wide on each side. Beyond each end, the nearest point of the segment is the end itself, so the points within 4 m of it make half a disc of radius 4 m around that end.
A wall along the segment cuts this shape in half, leaving the strip on one side and a quarter circle at each end.
Worked example: A Guard Dog on a Running Wire Along a Wall: the Ground It Covers, and What It Can Reach
Question A guard dog's lead, 4 m long, is clipped to a ring that slides along a straight wire. The wire is fixed along the foot of a long straight wall, and the dog stays in the yard on one side of the wall. On a plan in meters, the wall is the x-axis, the yard is the region y ≥ 0, and the wire runs from (6, 0) to (16, 0). (a) Find the area of the yard the dog can reach, in terms of π. (b) The owner wants the dog kept away from a trash can at P(19, 2) and a bird table at Q(11, 5). Find how far each of P and Q is from the nearest point of the wire, and say which of them the dog can reach.
1.Beside the wire, the points within 4 m of it form a rectangle 10 m long and 4 m deep: 10 × 4 = 40 m².
Beside the wire the dog reaches a rectangle 10 m by 4 m: 40 m². 2.Beyond each end of the wire, the points within 4 m of that end, on the yard side of the wall, form a quarter circle of radius 4 m. Together the two quarter circles make 2 × 14 × π × 42 = 8π m².
Beyond each end, a quarter circle of radius 4 m: 8π m² for the two. 3.(a) The dog can reach 40 + 8π ≈ 65.1 m² of the yard.
(a) The dog can reach 40 + 8π ≈ 65.1 m². 4.Q(11, 5) is beside the wire, since 11 is between 6 and 16, so its nearest point of the wire is straight below it, (11, 0), which is 5 m away.
Q(11, 5) is beside the wire: its nearest point of the wire is (11, 0), 5 m away. 5.P(19, 2) is beyond the end of the wire at (16, 0), since 19 > 16, so its nearest point of the wire is that end: √32 + 22 = √13 ≈ 3.61 m away.
P(19, 2) is beyond the end (16, 0): it is √13 ≈ 3.61 m from that end. 6.(b) √13 ≈ 3.61 m is less than 4 m, so the dog can reach the trash can; 5 m is more than 4 m, so it cannot reach the bird table.
(b) The dog can reach the trash can at P, but not the bird table at Q.
Answer: (a) 40 + 8π ≈ 65.1 m²; (b) the trash can is √13 ≈ 3.61 m from the wire, so the dog can reach it; the bird table is 5 m from the wire, so the dog cannot
Common mistakes
- Giving the distance from P as 2 m, its distance from the wall. The point (19, 0) straight below P is not on the wire, which stops at x = 16; the nearest point of the wire is its end.
- Adding a semicircle at each end, 16π m² in all, as if the dog could go behind the wall. The wall cuts each end circle in half again, leaving a quarter circle on the yard side.
Worked example: Four Sprinklers at the Corners of a Lawn: the Dry Ground, and the Reach That Waters All of It
Question A rectangular lawn is 12 m long and 8 m wide. A sprinkler at each of its four corners waters every point of the lawn within 4 m of it. (a) Find the area of the lawn that no sprinkler waters, in terms of π. (b) The four sprinklers are replaced by stronger ones, all with the same reach. Find the least reach that waters the whole lawn, and the point of the lawn that is reached last.
1.The lawn's corners are right angles, so each sprinkler waters a quarter circle of radius 4 m on the lawn: 14 × π × 42 = 4π m².
Each sprinkler waters a quarter circle of radius 4 m: 4π m². 2.Along the 8 m sides two sprinklers are 8 m apart and 4 + 4 = 8, so their quarter circles only touch; along the 12 m sides they are farther apart. No point is watered twice, and the four sprinklers water 4 × 4π = 16π m².
Along the 8 m sides the quarter circles only touch, so the four water 16π m². 3.(a) The dry area is 12 × 8 − 16π = 96 − 16π ≈ 45.7 m².
(a) The dry area is 96 − 16π ≈ 45.7 m². 4.Split the lawn into four rectangles 6 m by 4 m, one at each corner. A point in one of them is nearest that corner, and the point of the rectangle farthest from the corner is the rectangle's opposite corner, which is the center of the lawn.
In each 6 m by 4 m quarter of the lawn, the point farthest from its corner is the center. 5.(b) The center is √62 + 42 = √52 = 2√13 ≈ 7.21 m from every sprinkler, so the least reach is 2√13 ≈ 7.21 m, and the center of the lawn is the last point reached.
(b) The center is 2√13 ≈ 7.21 m from every corner: that is the least reach.
Answer: (a) 96 − 16π ≈ 45.7 m²; (b) 2√13 ≈ 7.21 m, and the last point reached is the center of the lawn
Common mistakes
- Giving the length of the diagonal, √122 + 82 ≈ 14.4 m, as the least reach. That is what ONE sprinkler at a corner would need; with four, each point only has to be within reach of its nearest sprinkler.
- Taking four whole circles, 4 × 16π = 64π ≈ 201 m², from the lawn, which leaves a negative area. Only a quarter of each circle lies on the lawn.