The Central Idea
A locus is the set of all points that satisfy a specified condition.
The condition may involve:
- distance from a point;
- distance from a line;
- equal distance from two objects;
- a fixed sum or difference of distances;
- an angle condition;
- an algebraic equation.
The central question is:
\[ \boxed{\text{Which points satisfy the condition, and only that condition?}} \]Locus Means a Complete Solution Set
A locus is not one convenient example.
It must contain:
- every point that satisfies the condition;
- no point that fails the condition.
This two-way requirement is what makes locus reasoning rigorous.
Condition to Set
Locus reasoning translates a condition into a geometric set.
Symbolically:
\[ \text{condition on }P \longrightarrow \{P:\text{condition is true}\}. \]Geometry then identifies the shape or structure of that set.
Set to Condition
The process also works in reverse.
Given a familiar geometric set, ask what property every point on it shares.
For example, every point \(P\) on a circle centered at \(O\) satisfies
\[ OP=r. \]Thus the circle is the locus determined by a fixed-distance condition.
The Simplest Distance Locus
Fix a point \(O\) and a positive distance \(r\).
The locus of all points \(P\) satisfying
\[ OP=r \]is a circle centered at \(O\) with radius \(r\).
\[ \boxed{\{P:OP=r\}=\text{circle}(O,r)}. \]Inside, On, and Outside a Circle
The same center and radius create three related distance conditions.
- \(OP
- \(OP=r\): points on the circle;
- \(OP>r\): points outside the circle.
Equality usually describes a boundary; inequalities describe regions.
Disks and Circles
The condition
\[ OP\le r \]describes the closed disk consisting of the circle and its interior.
The condition
\[ OPFixed Distance from a Line
Fix a line \(\ell\) and a positive distance \(d\).
The locus of points whose perpendicular distance from \(\ell\) is exactly \(d\) consists of two lines parallel to \(\ell\), one on each side.
Why Two Parallel Lines Appear
A point can lie distance \(d\) from \(\ell\) on either side of the line.
Therefore the condition
\[ d(P,\ell)=d \]ordinarily produces two parallel components.
Points within a Fixed Distance of a Line
The condition
\[ d(P,\ell)\le d \]describes an infinite strip centered on \(\ell\).
Its boundary consists of the two lines parallel to \(\ell\) at distance \(d\).
Equidistant from Two Points
Fix distinct points \(A\) and \(B\).
The locus of points \(P\) satisfying
\[ PA=PB \]is the perpendicular bisector of \(\overline{AB}\).
\[ \boxed{\{P:PA=PB\}=\text{perpendicular bisector of }\overline{AB}}. \]Why the Perpendicular Bisector Is the Locus
The perpendicular-bisector theorem gives one direction:
\[ P\text{ on the perpendicular bisector} \Rightarrow PA=PB. \]Its converse gives the other:
\[ PA=PB \Rightarrow P\text{ on the perpendicular bisector}. \]Together they establish exact equivalence.
Closer to One of Two Points
The perpendicular bisector separates the plane into two half-planes.
On one side,
\[ PAEquidistant from Two Intersecting Lines
Fix two intersecting lines \(\ell_1\) and \(\ell_2\).
The locus of points whose perpendicular distances from the lines are equal is the pair of angle-bisector lines.
\[ \boxed{ d(P,\ell_1)=d(P,\ell_2) \Rightarrow P\text{ lies on an angle bisector.} } \]Internal and External Angle-Bisector Loci
Intersecting lines create two bisector lines.
One bisects one pair of vertical angles, and the other bisects the supplementary pair.
Together they form the complete locus of equal line-distance.
Inside a Single Angle
If the problem restricts \(P\) to the interior of one angle, then the locus of equal distances from the two sides is the internal angle bisector only.
Equidistant from Two Parallel Lines
Suppose \(\ell_1\parallel\ell_2\).
The locus of points equidistant from both is the line parallel to both and exactly halfway between them.
Why There Is One Midline
If the distance between the parallel lines is \(d\), then every point on the middle parallel has distance
\[ \frac d2 \]from each boundary line.
Two Circles as Competing Distance Conditions
Suppose a point \(P\) must satisfy
\[ PA=r_1 \]and
\[ PB=r_2. \]The first condition places \(P\) on one circle. The second places \(P\) on another.
The possible solutions are their intersection points.
Intersections Solve Simultaneous Geometric Conditions
This is the basic logic of many construction problems:
\[ P\in L_1 \]and
\[ P\in L_2. \]Therefore,
\[ \boxed{P\in L_1\cap L_2}. \]Zero, One, or Two Intersections
Two circles may have:
- no common points;
- one common point when tangent;
- two common points when crossing.
Locus intersections naturally encode existence and multiplicity of solutions.
Circle and Line Intersections
A line and circle may also meet in:
- zero points;
- one point if tangent;
- two points if secant.
The number of intersections is the number of points satisfying both conditions simultaneously.
Two Lines as Loci
Two nonparallel lines intersect in one point.
Two distinct parallel lines have no common point.
Coincident lines share infinitely many points.
These are also intersection-count statements about solution sets.
Construction through Locus Intersection
A classical construction often solves a problem by identifying two loci whose intersection must contain the desired point.
The compass draws distance loci. The straightedge connects known points and creates line loci.
Example: Construct a Point Equidistant from \(A\) and \(B\)
The condition is
\[ PA=PB. \]Therefore \(P\) must lie on the perpendicular bisector of \(\overline{AB}\).
Any point on that line satisfies the requirement.
Example: Add a Fixed-Distance Condition
Now require
\[ PA=PB \]and
\[ PC=r \]for fixed point \(C\).
The first locus is a perpendicular bisector. The second is a circle.
Their intersections are exactly the desired points.
Example: Equal Distance from Two Sides and a Fixed Point
Suppose \(P\) must be inside \(\angle ABC\), equally distant from both sides, and a fixed distance \(r\) from \(Q\).
Then \(P\) lies on:
- the internal angle bisector of \(\angle ABC\);
- the circle centered at \(Q\) with radius \(r\).
Their intersections are the candidate solutions.
The Circumcenter as an Intersection of Loci
A circumcenter is equidistant from the three vertices of a triangle.
Therefore it lies on the perpendicular bisector of each side.
The intersection of two such bisectors determines the point; the third passes through the same location.
The Incenter as an Intersection of Loci
An incenter is equidistant from the three side lines of a triangle.
Therefore it lies on the internal angle bisector of each vertex angle.
Their common intersection is the incenter.
Locus of Points at a Fixed Sum of Distances
Fix points \(F_1\) and \(F_2\).
The locus of points \(P\) satisfying
\[ PF_1+PF_2=c \]for an appropriate positive constant \(c\) is an ellipse.
Why the Ellipse Is a Locus
Every point on the ellipse satisfies the same fixed sum of distances to the two foci.
Conversely, every point satisfying that fixed-sum condition belongs to the ellipse.
Locus of Points at a Fixed Difference of Distances
For fixed foci \(F_1\) and \(F_2\), the locus satisfying
\[ |PF_1-PF_2|=c \]for a suitable positive constant is a hyperbola.
The Absolute Value Matters
One branch may contain points for which
\[ PF_1-PF_2=c, \]while the other contains points for which
\[ PF_2-PF_1=c. \]The absolute value combines both branches.
Locus Equidistant from a Point and a Line
Fix a point \(F\) and a line \(\ell\).
The locus of points \(P\) satisfying
\[ PF=d(P,\ell) \]is a parabola.
Focus and Directrix
In the parabola definition:
- \(F\) is the focus;
- \(\ell\) is the directrix.
Every point on the parabola is equally distant from these two defining objects.
Conic Sections as Loci
Circle, ellipse, parabola, and hyperbola can all be understood through distance conditions.
This locus viewpoint unifies figures that may otherwise seem unrelated.
Coordinate Geometry Turns Loci into Equations
If a variable point has coordinates
\[ P(x,y), \]a geometric condition on \(P\) can often be translated into an equation in \(x\) and \(y\).
The graph of that equation is the locus.
Coordinate Circle Locus
Fix center
\[ O(h,k) \]and radius \(r\).
The condition
\[ OP=r \]becomes
\[ \boxed{(x-h)^2+(y-k)^2=r^2}. \]Coordinate Perpendicular-Bisector Locus
Let
\[ A(x_1,y_1),\qquad B(x_2,y_2). \]The locus condition
\[ PA=PB \]becomes
\[ (x-x_1)^2+(y-y_1)^2 = (x-x_2)^2+(y-y_2)^2. \]Why the Quadratic Terms Cancel
Expanding both squared-distance expressions produces \(x^2\) and \(y^2\) on both sides.
They cancel, leaving a linear equation.
This matches the geometric fact that the locus is a line.
Coordinate Example: Equal Distance from Two Points
Let
\[ A(0,0),\qquad B(8,0). \]Set the squared distances equal:
\[ x^2+y^2=(x-8)^2+y^2. \]Simplifying gives
\[ \boxed{x=4}. \]This is the perpendicular bisector of \(\overline{AB}\).
Coordinate Locus at Fixed Distance from an Axis
The distance from point \((x,y)\) to the \(x\)-axis is
\[ |y|. \]Therefore
\[ |y|=3 \]gives
\[ \boxed{y=3\text{ or }y=-3}. \]Coordinate Locus Equidistant from the Axes
Distance to the \(y\)-axis is
\[ |x|, \]while distance to the \(x\)-axis is
\[ |y|. \]Equal distance requires
\[ |x|=|y|. \]Hence
\[ \boxed{y=x\quad\text{or}\quad y=-x}. \]Coordinate Distance from a Point to a Line
For line
\[ Ax+By+C=0, \]the distance from \(P(x_0,y_0)\) to the line is
\[ \boxed{ d= \frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}. } \]Coordinate Angle-Bisector Locus
For lines
\[ A_1x+B_1y+C_1=0 \]and
\[ A_2x+B_2y+C_2=0, \]equal perpendicular distances give
\[ \frac{|A_1x+B_1y+C_1|}{\sqrt{A_1^2+B_1^2}} = \frac{|A_2x+B_2y+C_2|}{\sqrt{A_2^2+B_2^2}}. \]Removing the Absolute Values
The angle-bisector equations can be written
\[ \boxed{ \frac{A_1x+B_1y+C_1}{\sqrt{A_1^2+B_1^2}} = \pm \frac{A_2x+B_2y+C_2}{\sqrt{A_2^2+B_2^2}}. } \]The two signs correspond to the two angle-bisector lines.
Inequality Loci
Locus reasoning is not limited to equality.
Conditions such as
\[ PAdescribe regions rather than only boundary curves.
Boundary and Region
Equality often defines the boundary:
\[ PA=PB. \]An inequality chooses one side or region:
\[ PAIntersection of Inequality Loci
Multiple inequality conditions can create a feasible region.
For example:
\[ d(P,A)\le 5 \]and
\[ d(P,B)\le 4 \]describe the overlap of two closed disks.
Loci and Feasible Regions
A feasible region is the set of points satisfying all constraints simultaneously.
In geometric language:
\[ \boxed{\text{feasible region}=\bigcap \text{constraint loci}.} \]Loci in Optimization
Once constraints define an allowed locus or region, an optimization problem may ask which point minimizes or maximizes some quantity.
Locus reasoning therefore provides the geometric domain on which optimization operates.
Nearest-Point Reasoning
The nearest point on a line to an external point lies along the perpendicular.
Thus perpendicular constructions often arise when a locus problem asks for minimum distance.
Equal-Distance Boundaries in Networks
If several facilities compete by nearest distance, perpendicular bisectors between pairs of facilities form boundaries.
These boundaries combine into Voronoi regions.
Voronoi Cells
A Voronoi cell for site \(A\) consists of all points at least as close to \(A\) as to every competing site.
Its edges occur where two site distances are equal.
Voronoi Vertices
A Voronoi vertex is often equidistant from three sites.
Geometrically, it can occur where perpendicular bisectors of site pairs intersect.
This connects elementary locus reasoning to computational geometry.
Loci and Navigation
Locus conditions can describe possible positions in navigation.
If a receiver is known to be a fixed distance from one beacon, it lies on a circle centered at that beacon.
A second distance measurement creates another circle, and their intersections narrow the possible locations.
Two-Distance Positioning
Suppose
\[ PA=r_1 \]and
\[ PB=r_2. \]Two circle intersections generally leave two possible positions.
A third independent condition can remove the ambiguity.
Three-Distance Positioning
A third distance
\[ PC=r_3 \]adds another circle in a planar model.
A common intersection, when consistent, identifies the position.
This is the geometric foundation of trilateration.
Loci in Three Dimensions
The same conditions can produce different objects in higher dimensions.
In three-dimensional Euclidean space:
\[ OP=r \]describes a sphere rather than a circle.
Equidistant from Two Points in Three Dimensions
In three dimensions, the locus
\[ PA=PB \]is a plane: the perpendicular-bisector plane of \(\overline{AB}\).
Dimensional context matters.
Fixed Distance from a Plane
In three dimensions, points a fixed positive distance \(d\) from a plane lie on two parallel planes, one on each side.
This is the spatial analogue of two parallel lines at fixed distance from a line in the plane.
Locus Reasoning Depends on the Ambient Space
The same verbal condition may produce different-dimensional objects depending on whether we work on a line, in a plane, or in space.
Always identify the ambient space before naming the locus.
One-Dimensional Example
On a number line, points exactly distance \(3\) from \(0\) are
\[ \boxed{-3\text{ and }3}. \]In a plane, the analogous condition produces a circle.
In three dimensions, it produces a sphere.
Locus Reasoning and Symmetry
Many loci are symmetry sets.
The perpendicular bisector is the reflection axis exchanging \(A\) and \(B\).
An angle bisector is a reflection axis exchanging the sides of the angle.
Symmetry Explains Equal Distance
A point fixed by a reflection must have equal distances to objects exchanged by that reflection.
This gives a transformation-based explanation for several familiar locus theorems.
Locus Reasoning and Proof
A locus claim usually requires two proof directions.
- Show that every point on the proposed locus satisfies the condition.
- Show that every point satisfying the condition lies on the proposed locus.
Proving only one direction establishes containment, not full equality of sets.
Necessary versus Sufficient Conditions
If every point on set \(L\) satisfies condition \(C\), then membership in \(L\) is sufficient for \(C\).
If every point satisfying \(C\) lies on \(L\), then membership in \(L\) is necessary for \(C\).
A true locus characterization needs both.
Example: Only One Direction Is Not Enough
Suppose we prove that every point on line \(\ell\) is equidistant from \(A\) and \(B\).
This shows
\[ \ell\subseteq\{P:PA=PB\}. \]To prove that \(\ell\) is the entire locus, we must also prove that no equidistant point lies elsewhere.
Constructive Locus Proofs
A construction can sometimes help prove a locus theorem.
Equal-radius circles, perpendiculars, congruent triangles, and reflection arguments often show why every constructed point satisfies the defining condition.
Algebraic Locus Proofs
Coordinate methods can prove the same relationship by translating the condition into an equation and simplifying.
This gives a second representation of the same geometric solution set.
Geometric and Algebraic Descriptions Should Agree
If geometry predicts that a locus is a line, the resulting coordinate equation should simplify to a linear equation.
If geometry predicts a circle, the equation should have the corresponding quadratic form.
Disagreement is a warning that the setup or algebra may contain an error.
Parameterizing a Locus
Sometimes a locus is described not by one implicit equation but by a parameter.
For example, a circle may be written
\[ x=h+r\cos t, \] \[ y=k+r\sin t. \]As \(t\) varies, the point traces the entire circle.
Tracing versus Characterizing
A parametric rule tells how to generate points on a locus.
An implicit condition tells what property characterizes the locus.
Both viewpoints can describe the same geometric set.
Dynamic Geometry and Loci
Dynamic geometry software can move a point subject to constraints and trace the path of another dependent point.
The traced path suggests a locus.
A proof is then needed to establish the exact characterization.
Experimental Locus Discovery
A useful workflow is:
- construct the constrained configuration;
- drag a free point;
- trace the dependent point;
- identify the apparent curve;
- derive or prove the defining relationship.
Common Error: Giving Only One Point
A locus asks for all points satisfying the condition.
Producing one valid point does not identify the locus.
Common Error: Giving Only Part of the Locus
The set of points distance \(3\) from the \(x\)-axis is not just
\[ y=3. \]It is
\[ \boxed{y=3\text{ and }y=-3}. \]Both sides satisfy the condition.
Common Error: Forgetting an Interior Restriction
Equal distance from two intersecting lines gives two angle-bisector lines.
If the problem specifies points inside one chosen angle, only the internal bisector is relevant.
Common Error: Using Ordinary Distance to a Line
Distance from a point to a line means perpendicular distance.
A random segment from the point to the line does not define the line-distance used in locus problems.
Common Error: Confusing Boundary with Region
The equation
\[ OP=r \]describes the circle only.
The inequality
\[ OP\le r \]includes the entire disk.
Common Error: Ignoring Absolute Value
Distance is nonnegative.
Conditions involving difference of distances often require absolute value:
\[ |PF_1-PF_2|=c. \]Omitting it may discard one branch of the locus.
Common Error: Assuming Every Pair of Loci Intersects
Two geometric conditions may be incompatible.
Their loci may have no common points.
Always check existence rather than assuming a solution.
Common Error: Assuming One Intersection
Two loci may intersect in zero, one, two, finitely many, or infinitely many points.
The number of intersections depends on the geometric configuration.
Common Error: Proving Only One Implication
To establish that a set is exactly the locus, both directions are needed:
\[ P\in L\Rightarrow C(P) \]and
\[ C(P)\Rightarrow P\in L. \]Common Error: Ignoring Dimensional Context
Fixed distance from a point gives:
- two points on a line;
- a circle in a plane;
- a sphere in three-dimensional space.
The ambient space must be known.
Common Error: Trusting the Sketch Instead of the Condition
A locus may extend far beyond the portion shown in a diagram.
Infinite lines, curves, and regions should be interpreted from their mathematical conditions, not from the finite drawing window.
A Reliable Procedure for Locus Problems
- Identify the variable point.
- Write the condition that point must satisfy.
- Recognize any standard locus associated with the condition.
- Apply any domain, interior, side, or dimensional restrictions.
- If multiple conditions are present, intersect the individual loci.
- Check how many intersections or components result.
- Verify both directions if proving a full locus characterization.
A Coordinate Procedure for Locus Problems
- Assign coordinates to the variable point \(P(x,y)\).
- Translate each geometric condition into an equation or inequality.
- Simplify algebraically.
- Identify the resulting graph.
- Check for restrictions introduced by squaring, absolute values, or denominators.
- Compare the algebraic result with the expected geometry.
Example: Full Locus Workflow
Find all points \(P\) that are equidistant from
\[ A(-2,0) \]and
\[ B(6,0). \]The standard geometric locus is the perpendicular bisector of \(\overline{AB}\).
Its midpoint is
\[ (2,0), \]and the segment is horizontal, so the locus is
\[ \boxed{x=2}. \]Example: Two Conditions at Once
Find points \(P\) satisfying
\[ PA=PB \]and
\[ PC=5. \]First construct the perpendicular bisector of \(\overline{AB}\).
Then construct the circle centered at \(C\) with radius \(5\).
The locus solutions are the intersection points.
Example: Region Closer to One Point
Find the region satisfying
\[ PAThe desired locus is the open half-plane on the side containing \(A\).
Example: Fixed Distance from a Line
Find all points at distance \(4\) from line \(\ell\).
Construct two lines parallel to \(\ell\), one on each side, each exactly distance \(4\) from \(\ell\).
Their union is the complete locus.
Example: Equal Distance from Two Parallel Lines
Suppose parallel lines \(\ell_1\) and \(\ell_2\) are \(10\) units apart.
The equidistant locus is the parallel line halfway between them.
Every point on that line is
\[ \boxed{5} \]units from each original line.
Example: Equal Distance from Coordinate Axes
The condition is
\[ |x|=|y|. \]Therefore the locus consists of
\[ \boxed{y=x\quad\text{and}\quad y=-x}. \]Practice 1: Circle Locus
Point \(O\) is fixed. Describe the locus of all points \(P\) satisfying
\[ OP=7. \]Practice 2: Equal Distance from Two Points
Describe the locus of all points \(P\) satisfying
\[ PA=PB \]for distinct fixed points \(A\) and \(B\).
Practice 3: Distance from a Line
Describe the locus of all points exactly \(3\) units from a fixed line \(\ell\).
Practice 4: Equal Distance from Angle Sides
Point \(P\) is restricted to the interior of \(\angle ABC\) and is equally distant from its two sides.
Where must \(P\) lie?
Practice 5: Coordinate Locus
Find the locus of points \(P(x,y)\) satisfying
\[ |y|=5. \]Practice 6: Two Circle Conditions
A point \(P\) must satisfy
\[ PA=4 \]and
\[ PB=6. \]How should the possible locations of \(P\) be found geometrically?
Practice 7: Closer to \(A\)
Describe the region of points satisfying
\[ PAPractice 8: Conic Locus
What familiar curve is defined by
\[ PF_1+PF_2=c \]for two fixed foci and an appropriate constant \(c\)?
Practice 9: Three-Dimensional Locus
In three-dimensional space, describe the locus of points exactly \(r\) units from a fixed point \(O\).
Practice 10: Proof Structure
Why is proving only
\[ P\in L\Rightarrow C(P) \]insufficient to show that \(L\) is exactly the locus of condition \(C\)?
Practice Answers
Practice 1
The locus is a circle centered at \(O\) with radius \(7\):
\[ \boxed{\text{circle}(O,7)}. \]Practice 2
The locus is
\[ \boxed{\text{the perpendicular bisector of }\overline{AB}}. \]Practice 3
The locus consists of two lines parallel to \(\ell\), one on each side, each at perpendicular distance \(3\).
Practice 4
Since \(P\) is inside the angle and equally distant from its sides,
\[ \boxed{P\text{ lies on the internal angle bisector}.} \]Practice 5
The equation
\[ |y|=5 \]gives
\[ \boxed{y=5\text{ or }y=-5}. \]Practice 6
Construct:
- a circle centered at \(A\) with radius \(4\);
- a circle centered at \(B\) with radius \(6\).
Their intersection points, if any, are exactly the possible locations of \(P\).
Practice 7
The perpendicular bisector of \(\overline{AB}\) is the equality boundary.
The desired locus is the half-plane on the \(A\)-side of that boundary, excluding the boundary itself.
Practice 8
The fixed-sum-of-distances locus is
\[ \boxed{\text{an ellipse}}. \]Practice 9
In three dimensions, fixed distance \(r\) from \(O\) gives
\[ \boxed{\text{a sphere centered at }O\text{ with radius }r}. \]Practice 10
That implication shows only that every point on \(L\) satisfies the condition.
To prove that \(L\) is the complete locus, we must also show
\[ \boxed{C(P)\Rightarrow P\in L}. \]Otherwise additional points outside \(L\) might satisfy the condition.
What to Remember
Locus reasoning treats geometry as the study of complete solution sets to spatial conditions.
- A locus contains every point satisfying a condition and no points that fail it.
- Fixed distance from a point gives a circle in the plane.
- Equal distance from two points gives a perpendicular bisector.
- Equal distance from two intersecting lines gives angle bisectors.
- Fixed distance from a line gives two parallel lines.
- Multiple conditions are solved by intersecting their loci.
- Equality conditions often define boundaries, while inequalities define regions.
- Coordinate equations provide algebraic descriptions of geometric loci.
- Ellipse, parabola, and hyperbola can all be defined by distance conditions.
- A complete locus proof requires both directions of implication.
The essential habit is: translate each geometric condition into a complete set of possible points, then intersect those sets when several conditions must hold at once.
Where We Go Next
Locus reasoning identifies where geometric conditions can hold. The next step is to compare whole figures and determine when their shapes and dimensions are forced to match.
The next lesson, Congruence and Similarity Criteria, develops the conditions that establish exact sameness of size and shape or proportional sameness of shape, with particular attention to triangles and proof.