The Central Idea
A locus is the complete set of points satisfying a specified geometric condition.
The condition may involve:
- a fixed distance from a point;
- equal distances from two points;
- a fixed distance from a line;
- equal distances from two lines;
- distance from a segment, ray, circle, or other geometric object.
The essential habit is to translate a verbal distance condition into a geometric object or equation.
What Does Locus Mean?
The word locus means a place or position. In geometry, it refers to all positions that satisfy the same rule.
A locus is therefore not usually one isolated point. It may be:
- a line;
- a pair of lines;
- a circle;
- an arc;
- a curve;
- a region;
- several disconnected pieces;
- or, in special cases, a single point or no points at all.
Conditions Define the Locus
A geometric locus must contain every point satisfying the condition and only points satisfying the condition.
If a proposed figure contains extra points that do not satisfy the rule, it is too large. If it omits valid points, it is too small.
This two-direction requirement is important when proving a locus statement.
Distance from a Point
Suppose \(C\) is a fixed point and \(r>0\) is a fixed distance.
The locus of all points \(P\) satisfying
\[ CP=r \]is a circle centered at \(C\) with radius \(r\).
\[ \boxed{\text{fixed distance from a point}\longrightarrow\text{circle}} \]Why a Circle Is a Distance Locus
By definition, every point on a circle is the same distance from its center.
If \(P\) lies on the circle with center \(C\) and radius \(r\), then
\[ CP=r. \]Conversely, if a point \(P\) satisfies \(CP=r\), then \(P\) lies on that circle.
Both directions are satisfied, so the circle is exactly the required locus.
Distance Less Than or Equal to a Fixed Value
The locus changes when the condition changes.
If
\[ CP=r, \]the locus is the circle itself.
If instead
\[ CP\le r, \]the locus includes every point inside the circle as well as its boundary.
This filled region is the closed disk of radius \(r\).
Distance Greater Than a Fixed Value
If
\[ CP>r, \]the locus consists of all points outside the circle of radius \(r\), not including the circle itself.
Equality and inequality signs matter geometrically because they determine whether boundaries are included.
Points Equidistant from Two Fixed Points
Let \(A\) and \(B\) be distinct fixed points.
Consider every point \(P\) satisfying
\[ PA=PB. \]The locus is the perpendicular bisector of \(\overline{AB}\).
\[ \boxed{ PA=PB \longrightarrow P\text{ lies on the perpendicular bisector of }\overline{AB} }. \]The Perpendicular Bisector Theorem
The perpendicular bisector theorem states:
If a point lies on the perpendicular bisector of a segment, then it is equidistant from the endpoints of the segment.
If \(P\) lies on the perpendicular bisector of \(\overline{AB}\), then
\[ \boxed{PA=PB}. \]The Converse Perpendicular Bisector Theorem
The converse states:
If a point is equidistant from the endpoints of a segment, then it lies on the perpendicular bisector of that segment.
Thus,
\[ PA=PB \]implies that \(P\) lies on the perpendicular bisector of \(\overline{AB}\).
The theorem and its converse together establish the full locus statement.
Example: Equidistant from Two Points
Suppose \(A(-4,0)\) and \(B(4,0)\).
Their midpoint is
\[ M(0,0). \]Segment \(\overline{AB}\) is horizontal, so its perpendicular bisector is vertical.
Therefore, the locus of points equidistant from \(A\) and \(B\) is
\[ \boxed{x=0}. \]Algebraic Verification of an Equidistance Locus
Let \(P(x,y)\), \(A(-4,0)\), and \(B(4,0)\).
The condition
\[ PA=PB \]gives
\[ \sqrt{(x+4)^2+y^2} = \sqrt{(x-4)^2+y^2}. \]Squaring both sides:
\[ (x+4)^2+y^2=(x-4)^2+y^2. \]Cancel \(y^2\) and expand:
\[ x^2+8x+16=x^2-8x+16. \] \[ 16x=0. \] \[ \boxed{x=0}. \]The algebra reproduces the perpendicular-bisector result.
Distance from a Line
The distance from a point to a line is measured along the perpendicular segment from the point to the line.
This is the shortest possible distance from the point to any point on the line.
Therefore, distance-to-line conditions are fundamentally perpendicular distance conditions.
Points a Fixed Distance from a Line
Let line \(\ell\) be fixed, and let \(d>0\).
The locus of points exactly distance \(d\) from \(\ell\) consists of two lines parallel to \(\ell\), one on each side.
\[ \boxed{ \text{fixed distance from a line} \longrightarrow \text{two parallel lines} }. \]The existence of two components is easy to overlook.
Example: Fixed Distance from the \(x\)-Axis
The distance from \((x,y)\) to the \(x\)-axis is
\[ |y|. \]If the required distance is \(5\), then
\[ |y|=5. \]Therefore,
\[ \boxed{y=5\qquad\text{or}\qquad y=-5}. \]These are two lines parallel to the \(x\)-axis.
Points Equidistant from Two Parallel Lines
Suppose two distinct lines are parallel.
The locus of points equidistant from them is the line halfway between the two original lines and parallel to both.
For example, the lines
\[ y=2 \qquad\text{and}\qquad y=10 \]are eight units apart. Their equidistance locus is
\[ \boxed{y=6}. \]Points Equidistant from Two Intersecting Lines
If two lines intersect, the locus of points equidistant from the two lines consists of the angle bisectors of the angles formed by those lines.
Because two intersecting lines create two pairs of vertical angles, there are two perpendicular angle-bisector lines.
\[ \boxed{ \text{equidistant from two intersecting lines} \longrightarrow \text{their angle bisectors} }. \]The Angle Bisector Distance Theorem
A point on an angle bisector is equidistant from the two sides of the angle.
If \(P\) lies on the bisector of \(\angle ABC\), then the perpendicular distances from \(P\) to rays \(\overrightarrow{BA}\) and \(\overrightarrow{BC}\) are equal.
The distances must be measured perpendicularly to the sides.
The Converse Angle Bisector Distance Theorem
If a point in the interior of an angle is equidistant from the two sides of the angle, then the point lies on the angle bisector.
This converse allows equal perpendicular distances to be used as evidence that a point belongs to an angle bisector.
Intersection of Loci
Many geometry problems impose more than one condition.
In that case, find the locus for each condition separately and then locate their intersection.
For example:
- points \(4\) units from \(A\) form a circle centered at \(A\);
- points equidistant from \(B\) and \(C\) form the perpendicular bisector of \(\overline{BC}\);
- points satisfying both conditions are the intersections of that circle and line.
The Number of Solutions Can Vary
Two loci may intersect in:
- no points;
- one point;
- two points;
- several points;
- or infinitely many points if the loci overlap.
The geometry of the individual loci determines the number of simultaneous solutions.
Example: Circle and Perpendicular Bisector
Let \(A(-3,0)\) and \(B(3,0)\).
The locus equidistant from \(A\) and \(B\) is
\[ x=0. \]Suppose a point must also be \(5\) units from the origin.
That condition gives
\[ x^2+y^2=25. \]Set \(x=0\):
\[ y^2=25. \] \[ y=\pm5. \]Therefore, the two points satisfying both conditions are
\[ \boxed{(0,5)\qquad\text{and}\qquad(0,-5)}. \]The Circumcenter as an Intersection of Loci
In triangle \(ABC\), the perpendicular bisector of \(\overline{AB}\) is the locus of points equidistant from \(A\) and \(B\).
The perpendicular bisector of \(\overline{BC}\) is the locus of points equidistant from \(B\) and \(C\).
Their intersection is therefore equidistant from all three vertices:
\[ OA=OB=OC. \]This point \(O\) is the circumcenter of the triangle.
Why Two Perpendicular Bisectors Are Enough
If
\[ OA=OB \]and
\[ OB=OC, \]then by transitivity,
\[ OA=OC. \]Therefore, the intersection of two perpendicular bisectors automatically lies on the third perpendicular bisector as well.
The Incenter as an Intersection of Loci
The angle bisector of \(\angle A\) is the locus of interior points equidistant from sides \(\overline{AB}\) and \(\overline{AC}\).
A second angle bisector imposes another equal-distance condition.
Their intersection is equidistant from all three sides of the triangle.
This point is the incenter.
The common perpendicular distance from the incenter to the sides is the radius of the triangle's inscribed circle.
Loci and Geometric Constructions
Classical straightedge-and-compass constructions often work by building a locus and then finding an intersection.
For example:
- constructing a perpendicular bisector creates the locus of points equidistant from two endpoints;
- constructing an angle bisector creates the locus of points equidistant from two rays;
- drawing circles creates fixed-distance loci.
Construction procedures are therefore geometric implementations of distance conditions.
Constructing a Perpendicular Bisector
- Begin with segment \(\overline{AB}\).
- Choose a compass radius greater than half of \(AB\).
- Draw arcs centered at \(A\) above and below the segment.
- Without changing the radius, draw arcs centered at \(B\).
- Mark the two arc intersections.
- Draw the line through those intersections.
Each arc intersection is equidistant from \(A\) and \(B\), so both lie on the perpendicular bisector of \(\overline{AB}\).
Constructing an Angle Bisector
- Draw an arc centered at the angle's vertex so that it meets both sides.
- Use the same compass radius from the two intersection points to draw crossing arcs inside the angle.
- Connect the original vertex to the new arc intersection.
The resulting ray divides the angle into two congruent angles and forms the locus of interior points equidistant from the two sides.
Coordinate Distance Formula
For points
\[ A(x_1,y_1) \qquad\text{and}\qquad B(x_2,y_2), \]the Euclidean distance is
\[ \boxed{ AB= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} }. \]This formula converts geometric distance conditions into algebraic equations.
Using Squared Distance
When comparing two distances, square roots often cancel.
Instead of solving
\[ PA=PB, \]one may often compare squared distances:
\[ PA^2=PB^2. \]This avoids unnecessary radicals and usually simplifies coordinate locus equations.
Example: General Equidistance Equation
Let
\[ A(a_1,a_2), \qquad B(b_1,b_2), \qquad P(x,y). \]The condition \(PA=PB\) gives
\[ (x-a_1)^2+(y-a_2)^2 = (x-b_1)^2+(y-b_2)^2. \]The \(x^2\) and \(y^2\) terms cancel after expansion, leaving a linear equation.
This algebraic fact matches the geometric result: the equidistance locus of two distinct points is a line.
Distance from a Point to a Coordinate Line
For a horizontal line
\[ y=k, \]the distance from \((x,y)\) to the line is
\[ \boxed{|y-k|}. \]For a vertical line
\[ x=h, \]the distance is
\[ \boxed{|x-h|}. \]These formulas make many locus equations immediate.
Distance from a Point to a General Line
For the line
\[ Ax+By+C=0, \]the perpendicular distance from \(P(x_0,y_0)\) to the line is
\[ \boxed{ d= \frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}} }. \]This formula is a coordinate expression of perpendicular distance. It is especially useful when a locus is defined relative to oblique lines.
Example: Equidistant from the Coordinate Axes
The distance from \((x,y)\) to the \(x\)-axis is
\[ |y|, \]while the distance to the \(y\)-axis is
\[ |x|. \]Equidistance requires
\[ |x|=|y|. \]Therefore,
\[ \boxed{y=x\qquad\text{or}\qquad y=-x}. \]These are exactly the two angle bisectors formed by the coordinate axes.
Distance to a Segment Is More Subtle
The distance from a point to a segment is the shortest distance from the point to any point on that segment.
If the perpendicular projection lands inside the segment, the perpendicular length gives the distance.
If the perpendicular projection falls beyond an endpoint, the nearest endpoint determines the distance instead.
This distinction matters in more advanced loci involving finite objects.
Locus of Points a Fixed Distance from a Segment
The locus of points exactly distance \(r\) from a finite segment is not just two parallel segments.
It consists of:
- two segments parallel to the original segment at distance \(r\);
- joined at the ends by semicircular arcs of radius \(r\).
The resulting boundary has a rounded rectangular or “capsule” shape.
Regions Defined by Competing Distances
Locus reasoning also supports inequalities.
Suppose \(A\) and \(B\) are fixed points.
The perpendicular bisector separates the plane into two half-planes:
- one containing points closer to \(A\);
- the other containing points closer to \(B\);
- with the perpendicular bisector itself containing points equidistant from both.
Thus, a distance equality gives the boundary between two distance-inequality regions.
Example: Closer to One Point Than Another
Let
\[ A(-2,0) \qquad\text{and}\qquad B(2,0). \]Their perpendicular bisector is
\[ x=0. \]Points with \(x<0\) are closer to \(A\), while points with \(x>0\) are closer to \(B\).
Therefore,
\[ PAVoronoi Thinking
When several fixed points compete by distance, perpendicular bisectors partition the plane into regions.
Each region contains the points closer to one fixed point than to the others.
This idea leads to Voronoi diagrams, which appear in geography, facility placement, biology, computer graphics, and spatial optimization.
The full theory belongs to a later level, but the underlying geometry is already present in perpendicular-bisector loci.
Loci in Three Dimensions
The same distance ideas extend beyond the plane.
In three-dimensional Euclidean space:
- points a fixed distance from a point form a sphere;
- points equidistant from two points form a plane;
- points a fixed distance from a plane form two parallel planes.
Dimension changes the geometric object, but the defining distance condition remains the same.
Common Error: Giving Only Part of the Locus
The locus of points \(3\) units from a line consists of two parallel lines, not one.
A complete locus must include all valid points on every side of the reference object unless the problem imposes an additional restriction.
Common Error: Confusing a Circle with a Disk
The condition
\[ CP=r \]gives a circle.
The condition
\[ CP\le r \]gives the entire closed disk.
Equality describes a boundary; an inequality may describe a region.
Common Error: Measuring Distance to a Line Along a Slant
Point-to-line distance is always the perpendicular distance.
A longer oblique segment from the point to the line is not the distance between the point and the line.
Common Error: Using a Midpoint Without Perpendicularity
A line through the midpoint of \(\overline{AB}\) is not automatically the locus of points equidistant from \(A\) and \(B\).
It must be the perpendicular bisector.
Both midpoint and perpendicularity are essential.
Common Error: Forgetting the Converse
Showing that points on a perpendicular bisector are equidistant proves only one direction.
To establish that the perpendicular bisector is the complete locus, we also need the converse: every point equidistant from the endpoints lies on that perpendicular bisector.
A Reliable Procedure for Locus Problems
- Identify the fixed geometric objects.
- Translate the condition into a distance statement.
- Recall the standard locus associated with that condition.
- Include every component of the locus.
- If there are multiple conditions, find each locus separately.
- Intersect the loci to obtain simultaneous solutions.
- Use coordinates when algebraic verification is useful.
- Check whether equality or inequality changes the boundary or region.
- Verify that every proposed point satisfies the original condition.
A Core Locus Reference Table
| Condition | Locus in the Plane |
|---|---|
| Fixed distance from a point | Circle |
| Equidistant from two points | Perpendicular bisector |
| Fixed distance from a line | Two parallel lines |
| Equidistant from two parallel lines | Parallel line halfway between them |
| Equidistant from two intersecting lines | Two angle-bisector lines |
| Fixed distance from a segment | Two parallel pieces joined by semicircular ends |
Practice 1: Fixed Distance from a Point
Describe the locus of all points exactly \(7\) units from a fixed point \(A\).
Practice 2: Equidistant from Two Points
Describe the locus of points \(P\) satisfying
\[ PA=PB \]for distinct fixed points \(A\) and \(B\).
Practice 3: Fixed Distance from a Line
Describe the locus of points exactly \(4\) units from a fixed line \(\ell\).
Practice 4: Coordinate Locus
Find the locus of points exactly \(3\) units from the line
\[ y=5. \]Practice 5: Equidistant from Parallel Lines
Find the line containing all points equidistant from
\[ y=-2 \qquad\text{and}\qquad y=8. \]Practice 6: Equidistant from Two Intersecting Lines
Describe the locus of points equidistant from the \(x\)-axis and the \(y\)-axis.
Practice 7: Perpendicular Bisector Equation
Find the locus of points equidistant from
\[ A(-6,0) \qquad\text{and}\qquad B(2,0). \]Practice 8: Intersection of Loci
Find the points that are both:
- equidistant from \(A(-4,0)\) and \(B(4,0)\);
- and \(6\) units from the origin.
Practice 9: Circumcenter Reasoning
A point \(P\) lies on the perpendicular bisectors of \(\overline{AB}\) and \(\overline{BC}\).
What can be concluded about \(PA\), \(PB\), and \(PC\)?
Practice 10: Incenter Reasoning
A point lies on two interior angle bisectors of a triangle. What distance relationship does it have with the three sides?
Practice 11: Region by Distance
Let
\[ A(-5,0) \qquad\text{and}\qquad B(5,0). \]Which half-plane contains the points closer to \(A\) than to \(B\)?
Practice 12: Identify the Error
A student says:
“The locus of points \(2\) units from a horizontal line is one horizontal line \(2\) units above it.”
Explain the error.
Practice Answers
Practice 1
A fixed distance from a fixed point produces a circle.
\[ \boxed{\text{circle centered at }A\text{ with radius }7} \]Practice 2
Points equidistant from two fixed points form the perpendicular bisector of the segment joining them.
\[ \boxed{\text{perpendicular bisector of }\overline{AB}} \]Practice 3
There is one parallel locus line on each side of \(\ell\).
\[ \boxed{\text{two lines parallel to }\ell,\text{ each }4\text{ units away}} \]Practice 4
Distance \(3\) from \(y=5\) means
\[ |y-5|=3. \]Therefore,
\[ \boxed{y=8\qquad\text{or}\qquad y=2}. \]Practice 5
The halfway value between \(-2\) and \(8\) is
\[ \frac{-2+8}{2}=3. \]Therefore,
\[ \boxed{y=3}. \]Practice 6
Equidistance from the coordinate axes gives
\[ |x|=|y|. \]Therefore,
\[ \boxed{y=x\qquad\text{or}\qquad y=-x}. \]Practice 7
The midpoint of \((-6,0)\) and \((2,0)\) is
\[ (-2,0). \]The segment is horizontal, so its perpendicular bisector is vertical:
\[ \boxed{x=-2}. \]Practice 8
Equidistance from \((-4,0)\) and \((4,0)\) gives
\[ x=0. \]Distance \(6\) from the origin gives
\[ x^2+y^2=36. \]Substituting \(x=0\):
\[ y^2=36. \] \[ y=\pm6. \]Therefore,
\[ \boxed{(0,6)\qquad\text{and}\qquad(0,-6)}. \]Practice 9
From the first perpendicular bisector,
\[ PA=PB. \]From the second,
\[ PB=PC. \]Therefore,
\[ \boxed{PA=PB=PC}. \]Practice 10
The intersection of two interior angle bisectors is the incenter, which is equidistant from all three sides of the triangle.
\[ \boxed{\text{equal perpendicular distance from all three sides}} \]Practice 11
The perpendicular bisector of \(\overline{AB}\) is \(x=0\). Point \(A\) lies in the left half-plane.
Therefore, points closer to \(A\) satisfy
\[ \boxed{x<0}. \]Practice 12
The student included only one side of the original line. Points exactly \(2\) units below the line also satisfy the condition.
The complete locus is
\[ \boxed{\text{two parallel lines, one on each side}}. \]What to Remember
- a locus is the complete set of points satisfying a geometric condition;
- a fixed distance from a point produces a circle;
- points equidistant from two points lie on a perpendicular bisector;
- a fixed distance from a line produces two parallel lines;
- points equidistant from parallel lines lie halfway between them;
- points equidistant from intersecting lines lie on angle bisectors;
- multiple conditions are solved by intersecting their loci;
- coordinate distance formulas convert geometric locus conditions into equations;
- distance to a line is always measured perpendicularly;
- theorem-and-converse pairs are what establish a complete locus characterization.
The central translation is: distance condition \(\longleftrightarrow\) geometric set.
Where We Go Next
Locus problems focus on points and sets defined by distance. The final lesson in this Euclidean Relationships section broadens that reasoning to comparisons involving lengths, areas, volumes, and geometric constraints.
The next lesson, Area, Volume, and Geometric Inequalities, develops relationships that compare geometric quantities rather than requiring exact equality.