Geometry

Locus of points

You know the rule a point has to obey. Where are all the points that obey it?

You need

Segments, rays and lengthGrades 8 to 9 What a circle isMath 6

Where you are 8 in Points, lines and planes

Geometry11

Points, lines and planes8

▴ 3 above Angle types and pairs8 Lines and planes in space Parallel lines and transversals4 Segments, rays and length4

Locus of points

Why this exists

A pirate’s note says the gold is buried “ten paces from the old oak”. That does not tell you where to dig. It gives a whole ring of places. A second line, “and as far from the east post as from the west post”, gives a second set of places. The gold must be in both sets at once, so you dig only where they cross.

Geometry is full of questions shaped like this. You are not handed a point; you are handed a rule a point must obey, and asked where such points can be. The answer is rarely one point. It is a whole figure, and naming that figure exactly is the first step to using it. Even the circle is best understood this way, as every point one fixed distance from a center (what a circle is), and so are the parabola, the ellipse and the hyperbola.

Definition

A locus (plural loci) is the set of all the points that meet a stated condition, and only those points.

Both halves matter. To say that a figure is the locus you must know two things:

every point on the figure meets the condition, and
every point that meets the condition is on the figure.

A figure that fails the first has extra points in it. A figure that fails the second has missed some. A locus is also stated for a setting, usually a plane, sometimes all of space, and the same rule can give a different figure in each.

Intuition

Picture every point of the page testing itself against the rule and lighting up if it passes. The locus is the shape the lit points make.

Take the rule “distance r from the point O”. Points far from O stay dark, and so do points close to it. The ones that light up sit exactly r away, in every direction, and they draw a circle.

Figure 1Every point at distance r from O, and no other point, lies on this circle.

To find a locus you have not met before, do exactly this by hand: find four or five points that pass, sketch the shape they suggest, then check both halves of the definition.

Results

Five basic loci in a plane.

The points at distance r from a point O (with r > 0) form the circle with center O and radius r.
The points equidistant from two points A and B form the perpendicular bisector of \overline{AB}.
The points at distance d from a line \ell (with d > 0) form two lines parallel to \ell, one on each side, each at distance d from it.
The points equidistant from two parallel lines form the parallel line midway between them.
The points equidistant from two crossing lines form the two lines that bisect the angles at the crossing. Those two lines are perpendicular. Inside one of the four angles this reads: a point inside an angle is equally far from the lines of its two sides exactly when it lies on that angle’s bisector, the angle bisector distance theorem.

Figure 2Points as far from A as from B fill the perpendicular bisector of \overline{AB}.

Distance from a point to a line always means the shortest distance, measured along the perpendicular (distance from a point to a line).

Figure 3Points at distance d from \ell fill two lines, one on each side.

Figure 4Points as far from one crossing line as from the other fill the two angle bisectors.

Two conditions at once. The points meeting two conditions are the points common to the two loci: where they cross.

Why it is true

Result 1 is the definition of a circle, read in both directions.

For Result 2, put the points on a coordinate grid: A = (-a, 0) and B = (a, 0) with a > 0. The midpoint of \overline{AB} is the origin, and the perpendicular bisector is the line x = 0. For any point P = (x, y), the distance formula gives

\begin{aligned} PA^2 &= (x + a)^2 + y^2 \\ PB^2 &= (x - a)^2 + y^2 \\ PA^2 - PB^2 &= (x + a)^2 - (x - a)^2 = 4ax \end{aligned}

Distances are never negative, so PA = PB exactly when PA^2 = PB^2, that is, when 4ax = 0, that is, when x = 0. The word “exactly” carries both halves of the definition: every point on the line x = 0 is equidistant, and no point off it is. The same fact proved with congruent triangles is the perpendicular bisector theorem.

For Result 3, take \ell to be the x-axis. The perpendicular from (x, y) to it is vertical, so the distance is |y|, and |y| = d exactly when y = d or y = -d: two lines, one each side. For Result 4, with lines y = p and y = q, the distances |y - p| and |y - q| are equal exactly when \displaystyle y = \frac{p + q}{2}.

For Result 5, let the lines cross at V. The point V is at distance 0 from both lines and lies on every bisector, so take any other point P and drop perpendiculars to the two lines, with feet D and E.

Suppose PD = PE. Neither foot is V: if D = V, then \overline{PV} is perpendicular to the first line, so it is not perpendicular to the second, E \ne V, and PV is the hypotenuse of the right triangle PEV, longer than its leg PE = PD = PV, which is impossible. So \triangle PDV and \triangle PEV are right triangles with the same hypotenuse \overline{PV} and equal legs, and they are congruent by hypotenuse-leg. Hence \angle PVD = \angle PVE. The rays \overrightarrow{VD} and \overrightarrow{VE} lie on different lines, so they cannot sit on the same side of \overrightarrow{VP} at the same angle; \overrightarrow{VP} runs between them and bisects \angle DVE, which is one of the four angles at V.

Conversely, let P lie on the bisector of one of the four angles, each half measuring \theta. Since \theta < 90^\circ, each foot lands on a side of that angle, so \triangle PDV and \triangle PEV have right angles at D and E, equal angles \theta at V, and the common side \overline{PV}. They are congruent by angle-angle-side, and PD = PE.

So the locus is made of the bisectors of the four angles. Extend the bisector of one of them backwards through V: by vertical angles it cuts the opposite angle into the same two halves, so the four bisecting half-lines join into two whole lines. Two neighboring angles add to 180^\circ, so their halves add to 90^\circ, and the two lines are perpendicular.

Worked examples

Example 1

Exactly, at most, more than

In a plane, describe each set of points.

(a) The points exactly 5 cm from a point Q.

(b) The points at most 5 cm from Q.

Solution

(a) By Result 1, the circle with center Q and radius 5 cm.

(b) “At most 5” allows every distance from 0 up to 5. The points exactly 5 away form the circle, and the closer ones fill its inside, including Q itself. The locus is a solid disc, edge included.

(c) The points outside the circle. The circle itself is not included: a point on it is exactly 5 cm away, which is not more than 5.

So a locus need not be a thin curve. “At most” and “more than” give regions, and the exact words decide whether the edge belongs.

Example 2

Where two loci cross

Points A and B are 6 units apart. Find every point that is 4 units from A and is as far from A as from B.

Figure 5Two conditions, two loci. The points meeting both are where the loci cross.

Solution

Take one condition at a time. The points 4 from A form a circle with center A and radius 4 (Result 1). The points as far from A as from B form the perpendicular bisector of \overline{AB} (Result 2). The points we want lie on both.

The bisector crosses \overline{AB} at its midpoint M, with AM = 3. Since 3 < 4, the line passes through the inside of the circle, so it meets the circle twice, at P and Q.

To say where, use the triangle AMP. It has a right angle at M, leg AM = 3 and hypotenuse AP = 4, so by the Pythagorean theorem

MP = \sqrt{4^2 - 3^2} = \sqrt{7} \approx 2.65

So P and Q lie on the bisector, \displaystyle \sqrt{7} units either side of M.

The count depends on the radius. With radius 3 the circle would just touch the bisector, at M: one point. With radius 2 it would not reach it: no points.

Example 3

The equation of a locus

Find an equation for the set of points equidistant from A = (2, -1) and B = (6, 5).

Solution

Let P = (x, y). The condition is PA = PB. Neither distance is negative, so this says the same thing as PA^2 = PB^2, which needs no square roots. By the distance formula,

\begin{aligned} PA^2 &= (x - 2)^2 + (y + 1)^2 \\ &= x^2 + y^2 - 4x + 2y + 5 \\ PB^2 &= (x - 6)^2 + (y - 5)^2 \\ &= x^2 + y^2 - 12x - 10y + 61 \end{aligned}

Set them equal. The x^2 and y^2 appear on both sides and cancel, which is why the answer is a line:

\begin{aligned} -4x + 2y + 5 &= -12x - 10y + 61 \\ 8x + 12y &= 56 \\ 2x + 3y &= 14 \end{aligned}

The locus is the line 2x + 3y = 14.

Check. By Result 2 this should be the perpendicular bisector of \overline{AB}. The midpoint (4, 2) is on it, since 2(4) + 3(2) = 14. The line has slope \displaystyle -\frac{2}{3} and \overline{AB} has slope \displaystyle \frac{6}{4} = \frac{3}{2}; the product is -1, so they are perpendicular. Finding this line directly is its own skill, the equation of a perpendicular bisector.

Example 4

A line and a circle together

Find every point that is 3 units from the line y = 1 and 5 units from the origin.

Solution

The line y = 1 is horizontal, so the distance from (x, y) to it is measured straight up or down and equals |y - 1|. Then |y - 1| = 3 gives y = 4 or y = -2: two lines, as Result 3 says.

The points 5 from the origin form the circle x^2 + y^2 = 25. Cross it with each line in turn.

On y = 4: \;x^2 + 16 = 25, so x^2 = 9 and x = \pm 3.

On y = -2: \;x^2 + 4 = 25, so x^2 = 21 and \displaystyle x = \pm\sqrt{21}.

There are four points: (3, 4), (-3, 4), \displaystyle (\sqrt{21}, -2) and \displaystyle (-\sqrt{21}, -2). Each of the two lines meets the circle twice.

Example 5

A point and a line make a parabola

Find an equation for the set of points as far from the point F = (0, 1) as from the line y = -1.

Figure 6Every point as far from F as from the line y = -1 lies on one parabola.

Solution

Let P = (x, y). Its distance to F is \displaystyle \sqrt{x^2 + (y - 1)^2}. The line y = -1 is horizontal, so the distance from P to it is |y + 1|. Set them equal and square, which is allowed because both are distances:

\begin{aligned} x^2 + (y - 1)^2 &= (y + 1)^2 \\ x^2 + y^2 - 2y + 1 &= y^2 + 2y + 1 \\ x^2 &= 4y \end{aligned}

The locus is the curve x^2 = 4y, a parabola opening upward with its lowest point at the origin, halfway between F and the line.

Check. P = (4, 4) is on the curve, since 4^2 = 4(4). Then \displaystyle PF = \sqrt{4^2 + 3^2} = 5, and the distance from P down to y = -1 is 4 + 1 = 5.

This is no accident of the numbers. Every parabola is the locus of points equally far from one point and one line, which is how its focus and directrix are defined.

Boundaries

The setting changes the answer. In a plane, the points 3 cm from O form a circle; in space they form a sphere. In a plane, the points equidistant from A and B form a line; in space they form a plane, cutting \overline{AB} at its midpoint at right angles.

Distance to a line is not distance to a segment. Beside a segment the two agree. Past its ends, the nearest point of the segment is an endpoint, so the locus bends round in a half circle instead of running on straight.

A locus can be almost anything, including nothing. A curve, two separate lines, a region, a few points, one point, or empty. The points at distance 0 from O are just O. No point is at distance -2 from anything. Two loci may never cross, and then no point meets both conditions.

The figure must be exactly right. The midpoint of \overline{AB} is equidistant from A and B but is not the locus, since it misses the rest of the bisector. And a figure can have too much: the points P where \angle APB is a right angle lie on the circle with diameter \overline{AB}, but A and B are not in the locus, because at P = A there is no angle APB.

Where this goes next

Loci are how the conic sections are defined once you stop slicing cones. The circle keeps a constant distance from one point, the ellipse a constant sum of distances from two points, the hyperbola a constant difference (larger minus smaller), and the parabola equal distances from a point and a line, as in Example 5. Writing the condition in coordinates is how each equation is found, starting with the equation of a circle.

Crossing loci is also how a triangle gives up its special points. The point equally far from all three corners lies on all three perpendicular bisectors: the circumcenter. The point inside the triangle equally far from all three sides lies on the angle bisectors: the incenter.

And every arc a compass draws is a locus, the points at one distance from where the needle sits. That is why constructing a perpendicular bisector works.

Exercises

Describe the locus

In a plane, describe each locus in words.

The points 7 cm from a point C.
The points equidistant from the endpoints of a segment \overline{PQ}.
The points 2 cm from a line \ell.
The points equidistant from two parallel lines that are 10 cm apart.
The points at most 4 cm from a point D.
The points equidistant from two lines that cross at a point V.

Loci in coordinates

Find every point that is 13 units from the origin and lies on the line y = 5.
Find an equation for the points equidistant from (0, 0) and (6, 0).
Find an equation for the points equidistant from (1, 2) and (5, 6).
Find equations for the points 4 units from the line x = 2.
Find every point on the circle x^2 + y^2 = 25 that is 1 unit from the line y = 2.
Find every point that is 5 units from (0, 0) and 5 units from (6, 0).
Find an equation for the points as far from (0, 2) as from the line y = -2.

Two conditions in a plane

Points A and B are 10 cm apart. How many points are r cm from A and also r cm from B when (a) r = 6, (b) r = 5, (c) r = 4?
In Exercise 14 with r = 6, how far is each of the points from the line AB?
Points A and B are 8 cm apart. How many points are equidistant from A and B and 3 cm from the line AB, and how far is each of them from A?
A point P is 3 cm from a line \ell. How many points are 1 cm from \ell and 2 cm from P?

Reasoning

Explain why the centers of all circles of radius 3 cm that pass through a point A form a circle.
Three points A, B and C do not lie on one line. Explain, using loci, why exactly one point is equidistant from all three.
Two lines m and n cross. Explain why exactly four points are 2 cm from m and 2 cm from n, however large or small the angle between the lines.

Answers

The circle with center C and radius 7 cm.
The perpendicular bisector of \overline{PQ}.
Two lines parallel to \ell, one on each side, each 2 cm from it.
The line parallel to both and midway between them, 5 cm from each, since 10 \div 2 = 5.
The circle with center D and radius 4 cm and every point inside it: a solid disc, edge included.
The two lines through V that bisect the four angles at the crossing. They are perpendicular.
(12, 5) and (-12, 5). The point is (x, 5), so x^2 + 25 = 169 and x^2 = 144.
x = 3. Squaring PA = PB gives x^2 + y^2 = (x - 6)^2 + y^2, so 12x = 36.
x + y = 7. In (x - 1)^2 + (y - 2)^2 = (x - 5)^2 + (y - 6)^2 the squares cancel, leaving 8x + 8y = 56. The midpoint (3, 4) satisfies it.
x = 6 and x = -2. The distance to the vertical line x = 2 is |x - 2|, and |x - 2| = 4 gives x - 2 = \pm 4.
(4, 3), (-4, 3), \displaystyle (2\sqrt{6}, 1) and \displaystyle (-2\sqrt{6}, 1). The points 1 from y = 2 lie on y = 3 and y = 1. Then x^2 = 25 - 9 = 16, or x^2 = 25 - 1 = 24 and \displaystyle \sqrt{24} = 2\sqrt{6}.
(3, 4) and (3, -4). Equidistant from the two points means x = 3 (Exercise 8); then 9 + y^2 = 25.
x^2 = 8y. Squaring \displaystyle \sqrt{x^2 + (y - 2)^2} = |y + 2| gives x^2 + y^2 - 4y + 4 = y^2 + 4y + 4.
(a) 2, (b) 1, (c) 0. Such a point is equidistant from A and B, so it lies on the perpendicular bisector, 5 cm from A. The circle of radius r about A crosses that line twice if r > 5, touches it if r = 5, and misses it if r < 5.
\displaystyle \sqrt{11} cm, about 3.32 cm. Each point, A and the midpoint make a right triangle with hypotenuse 6 and one leg 5, so the other leg is \displaystyle \sqrt{36 - 25}.
Two points, each 5 cm from A. The perpendicular bisector of \overline{AB} crosses each of the two lines 3 cm from \overleftrightarrow{AB} once, 3 cm from the midpoint M on either side. With AM = 4, the right triangle gives \displaystyle \sqrt{4^2 + 3^2} = 5.
One point. The points 1 cm from \ell lie on two lines, 2 cm and 4 cm from P. The circle of radius 2 about P just touches the nearer line and cannot reach the farther one.
A circle of radius 3 passes through A exactly when its center is 3 cm from A. So the centers are the points 3 cm from A, which by Result 1 form the circle with center A and radius 3 cm.
A point equidistant from A and B lies on the perpendicular bisector of \overline{AB}, and one equidistant from B and C lies on that of \overline{BC}. As \overline{AB} and \overline{BC} are not parallel, neither are their bisectors, so these cross at exactly one point, which is then equidistant from all three: the circumcenter of triangle ABC.
The points 2 cm from m form two lines parallel to m, and those 2 cm from n form two lines parallel to n. Since m and n cross, no line of the first pair is parallel to one of the second, so each of the 2 meets each of the other 2 once: 2 \times 2 = 4 crossings. They are four different points, because the two lines of one pair never meet each other. The angle moves the points but never changes how many there are.