The Central Idea
A locus is the set of all points satisfying a specified geometric condition.
Coordinate geometry translates that condition into algebra.
The basic pattern is
\[ \boxed{ \text{geometric condition} \longleftrightarrow \text{equation or inequality} }. \]Solving or interpreting the algebraic condition reveals the corresponding geometric set.
Equations Describe Boundaries and Curves
An equation such as
\[ y=2x+3 \]describes all points lying on one line.
Likewise,
\[ x^2+y^2=25 \]describes all points lying on a circle of radius \(5\) centered at the origin.
Equations typically identify exact geometric boundaries or loci.
Inequalities Describe Regions
Inequalities often describe areas or volumes rather than only their boundaries.
For example,
\[ x^2+y^2<25 \]describes the interior of the circle of radius \(5\).
The inequality
\[ x^2+y^2\le25 \]includes both the interior and the circle itself.
Equality Versus Inequality
Compare
\[ x^2+y^2=9, \] \[ x^2+y^2<9, \]and
\[ x^2+y^2\le9. \]These represent, respectively:
- the circle of radius \(3\);
- the open disk inside the circle;
- the closed disk consisting of the circle and its interior.
A Fixed Distance from a Point
Suppose a point
\[ P=(x,y) \]must remain exactly \(r\) units from
\[ C=(h,k). \]The distance condition is
\[ \sqrt{(x-h)^2+(y-k)^2}=r. \]Squaring gives
\[ \boxed{(x-h)^2+(y-k)^2=r^2}. \]The locus is a circle.
Points Within a Fixed Distance
If a point must lie less than \(r\) units from
\[ (h,k), \]then
\[ \sqrt{(x-h)^2+(y-k)^2}Points At Most a Fixed Distance Away
The phrase at most \(r\) units away includes the boundary.
Therefore,
\[ \boxed{(x-h)^2+(y-k)^2\le r^2}. \]This describes a closed disk.
Points At Least a Fixed Distance Away
If a point must remain at least \(r\) units from
\[ (h,k), \]then
\[ \boxed{(x-h)^2+(y-k)^2\ge r^2}. \]The locus consists of the circle together with all points outside it.
Equidistant from Two Points
Let
\[ A=(x_1,y_1) \]and
\[ B=(x_2,y_2). \]A point
\[ P=(x,y) \]is equidistant from \(A\) and \(B\) when
\[ PA=PB. \]Algebraically,
\[ \sqrt{(x-x_1)^2+(y-y_1)^2} = \sqrt{(x-x_2)^2+(y-y_2)^2}. \]The Perpendicular-Bisector Locus
Squaring the equidistance equation gives
\[ (x-x_1)^2+(y-y_1)^2 = (x-x_2)^2+(y-y_2)^2. \]After expansion, the quadratic terms cancel.
The resulting equation is linear.
Therefore the locus of points equidistant from two distinct points is the perpendicular bisector of the segment joining them.
Example: Equidistant from Two Points
Find the locus of points equidistant from
\[ A=(-2,0) \]and
\[ B=(4,0). \]Set the squared distances equal:
\[ (x+2)^2+y^2 = (x-4)^2+y^2. \]Cancel \(y^2\) and expand:
\[ x^2+4x+4=x^2-8x+16. \]Therefore,
\[ 12x=12, \]so
\[ \boxed{x=1}. \]This is the perpendicular bisector of \(\overline{AB}\).
Closer to One Point Than Another
Suppose we want all points closer to
\[ A \]than to
\[ B. \]The condition is
\[ PAThis describes one side of the perpendicular bisector.
Half-Planes
A line divides the plane into two half-planes.
For example,
\[ y>2x+1 \]describes the open half-plane above the line
\[ y=2x+1. \]The inequality
\[ y\ge2x+1 \]includes the boundary line.
Testing Which Side of a Line
For a line in standard form
\[ Ax+By=C, \]the inequalities
\[ Ax+Bydescribe the two open half-planes.
A convenient test point can be substituted to determine which geometric side corresponds to each inequality.
Example: A Half-Plane
Consider
\[ 2x+y<6. \]The boundary is
\[ 2x+y=6. \]Test the origin:
\[ 2(0)+0<6. \]The statement is true, so the required half-plane is the side containing the origin.
Strips Between Parallel Lines
A region between two parallel lines can be expressed using a compound inequality.
For example,
\[ \boxed{2and
\[ x=7. \]Closed and Half-Closed Strips
The inequality
\[ 2\le x\le7 \]includes both boundary lines.
The inequality
\[ 2Algebraic inequality symbols encode boundary inclusion directly.
Distance from an Axis
The distance from
\[ (x,y) \]to the \(x\)-axis is
\[ |y|. \]Therefore points exactly \(3\) units from the \(x\)-axis satisfy
\[ |y|=3. \]Equivalently,
\[ \boxed{y=3\quad\text{or}\quad y=-3}. \]Distance from the \(y\)-Axis
The distance from
\[ (x,y) \]to the \(y\)-axis is
\[ |x|. \]Therefore points at most \(4\) units from the \(y\)-axis satisfy
\[ \boxed{|x|\le4}. \]This describes the closed vertical strip
\[ -4\le x\le4. \]Distance from a Horizontal Line
The distance from
\[ (x,y) \]to the horizontal line
\[ y=k \]is
\[ \boxed{|y-k|}. \]Thus points exactly \(d\) units from that line satisfy
\[ |y-k|=d. \]Distance from a Vertical Line
The distance from
\[ (x,y) \]to
\[ x=h \]is
\[ \boxed{|x-h|}. \]Points within \(d\) units of the line satisfy
\[ \boxed{|x-h|Distance from a General Line
The distance from
\[ (x_0,y_0) \]to the line
\[ Ax+By+C=0 \]is
\[ \boxed{ d= \frac{|Ax_0+By_0+C|} {\sqrt{A^2+B^2}} }. \]This formula can be used to construct loci at a fixed perpendicular distance from a line.
Lines Parallel to a Given Line at Fixed Distance
Points exactly \(d\) units from
\[ Ax+By+C=0 \]satisfy
\[ \frac{|Ax+By+C|} {\sqrt{A^2+B^2}} =d. \]Therefore,
\[ |Ax+By+C| = d\sqrt{A^2+B^2}. \]This splits into two parallel lines, one on each side of the original line.
Parabola as a Locus
A parabola is the set of points equidistant from a fixed point called the focus and a fixed line called the directrix.
Suppose the focus is
\[ (0,p) \]and the directrix is
\[ y=-p. \]A point
\[ (x,y) \]on the parabola satisfies
\[ \sqrt{x^2+(y-p)^2}=|y+p|. \]Deriving the Standard Parabola Equation
Square
\[ \sqrt{x^2+(y-p)^2}=|y+p|. \]Then
\[ x^2+(y-p)^2=(y+p)^2. \]Expand:
\[ x^2+y^2-2py+p^2 = y^2+2py+p^2. \]Therefore,
\[ \boxed{x^2=4py}. \]Ellipse as a Locus
An ellipse is the set of points for which the sum of distances to two fixed foci is constant.
If the foci are
\[ F_1,\qquad F_2, \]then an ellipse satisfies
\[ \boxed{PF_1+PF_2=2a}. \]The constant \(2a\) is greater than the distance between the foci.
Hyperbola as a Locus
A hyperbola is the set of points for which the absolute difference of distances to two fixed foci is constant.
Algebraically,
\[ \boxed{|PF_1-PF_2|=2a}. \]The absolute value allows the two branches to be described by one condition.
Angle Bisectors as Equidistance Loci
Points on an angle bisector are equidistant from the two sides of the angle.
In coordinate form, this can be expressed by setting the point-to-line distances equal.
Thus an angle-bisector construction can be converted into an algebraic distance equation.
Collinearity as an Algebraic Condition
Three points are collinear when they lie on the same line.
For nonvertical configurations, equal slopes provide one test:
\[ \frac{y_2-y_1}{x_2-x_1} = \frac{y_3-y_1}{x_3-x_1}. \]Cross-multiplying gives a form that also avoids division:
\[ \boxed{ (y_2-y_1)(x_3-x_1) = (y_3-y_1)(x_2-x_1) }. \]Collinearity and Determinants
Three planar points
\[ (x_1,y_1),\quad (x_2,y_2),\quad (x_3,y_3) \]are collinear when
\[ \boxed{ \begin{vmatrix} x_1&y_1&1\\ x_2&y_2&1\\ x_3&y_3&1 \end{vmatrix} =0 }. \]This determinant condition is a compact algebraic description of zero triangle area.
Parallelism as an Algebraic Condition
Two nonvertical lines are parallel when their slopes are equal.
In standard form,
\[ A_1x+B_1y=C_1 \]and
\[ A_2x+B_2y=C_2, \]their direction is parallel when the normal vectors are scalar multiples.
One test is
\[ \boxed{A_1B_2-A_2B_1=0}. \]Perpendicularity as an Algebraic Condition
For nonvertical lines with nonzero slopes,
\[ \boxed{m_1m_2=-1}. \]Using normal or direction vectors, perpendicularity can also be tested with a dot product equal to zero.
Thus a right-angle condition becomes an algebraic orthogonality condition.
Symmetry About the \(x\)-Axis
A locus is symmetric about the \(x\)-axis when replacing
\[ y \]with
\[ -y \]leaves its equation unchanged or equivalent.
For example,
\[ y^2=x \]is unchanged under
\[ y\mapsto -y. \]Symmetry About the \(y\)-Axis
Replace
\[ x \]with
\[ -x. \]If the equation remains equivalent, the locus is symmetric about the \(y\)-axis.
For example,
\[ y=x^2 \]has \(y\)-axis symmetry.
Symmetry About the Origin
Replace
\[ (x,y) \]with
\[ (-x,-y). \]If the equation remains equivalent, the locus is symmetric through the origin.
For example,
\[ y=x^3 \]has origin symmetry.
Intersections as Simultaneous Conditions
If a point must lie on two loci at once, it must satisfy both equations.
For example, intersection points of
\[ y=x+1 \]and
\[ x^2+y^2=25 \]are found by solving the system simultaneously.
Geometric intersection is therefore equivalent to shared algebraic solutions.
Unions of Loci
Sometimes a locus consists of points satisfying one condition or another.
For example,
\[ xy=0 \]factors as
\[ x=0 \]or
\[ y=0. \]Therefore the locus is the union of the two coordinate axes.
Products Can Encode Multiple Components
The equation
\[ (x-2)(y+3)=0 \]means
\[ x=2 \]or
\[ y=-3. \]The locus is the union of one vertical line and one horizontal line.
Absolute Value Can Encode Symmetric Loci
The equation
\[ |x|+|y|=4 \]describes all points whose coordinate magnitudes sum to \(4\).
The resulting locus is a square rotated \(45^\circ\) relative to the coordinate axes.
Absolute values naturally encode reflectional symmetry.
Example: Region from Absolute Values
The inequality
\[ |x|+|y|\le4 \]describes the interior and boundary of that rotated square.
Replacing equality by inequality changes a one-dimensional boundary into a two-dimensional region.
Loci in Three Dimensions
The same logic extends from \(\mathbb{R}^2\) to \(\mathbb{R}^3\).
For example,
\[ x^2+y^2+z^2=25 \]describes all points exactly \(5\) units from the origin.
The locus is a sphere.
Regions in Three Dimensions
The inequality
\[ x^2+y^2+z^2\le25 \]describes the solid ball of radius \(5\) centered at the origin.
Again, equality gives the boundary, while inequality gives a region.
Equidistant from Two Points in Space
In three dimensions, the locus of points equidistant from two distinct points is a plane rather than a line.
Algebraically,
\[ (x-x_1)^2+(y-y_1)^2+(z-z_1)^2 = (x-x_2)^2+(y-y_2)^2+(z-z_2)^2. \]The quadratic terms cancel, leaving a linear plane equation.
Perpendicular-Bisector Plane
The three-dimensional equidistance locus is the plane perpendicular to the segment joining the two fixed points and passing through its midpoint.
This is the spatial analogue of the perpendicular bisector in the plane.
Distance from a Coordinate Plane
The distance from
\[ (x,y,z) \]to the \(xy\)-plane is
\[ |z|. \]Therefore points exactly \(4\) units from the \(xy\)-plane satisfy
\[ |z|=4, \]or
\[ \boxed{z=4\quad\text{or}\quad z=-4}. \]Distance from the Origin in Higher Dimensions
In
\[ \mathbb{R}^n, \]the locus of points a fixed distance \(r\) from the origin satisfies
\[ \boxed{ x_1^2+x_2^2+\cdots+x_n^2=r^2 }. \]This is an \(n\)-dimensional sphere, often called a hypersphere.
Boundary, Interior, and Exterior
Algebraic descriptions often separate naturally into:
- boundary: equality;
- interior: strict inequality in one direction;
- exterior: strict inequality in the other direction.
For a circle,
\[ F(x,y)=x^2+y^2-r^2, \]the sign of \(F\) distinguishes inside, on, and outside.
Sign Functions as Region Classifiers
Suppose
\[ F(x,y)=Ax+By-C. \]Then:
- \(F(x,y)=0\) describes the boundary line;
- \(F(x,y)>0\) describes one side;
- \(F(x,y)<0\) describes the other side.
This viewpoint generalizes naturally to more complicated boundaries.
Constraints Can Be Combined
A geometric region may be defined by several simultaneous conditions.
For example,
\[ x^2+y^2\le25 \]together with
\[ y\ge0 \]describes the closed upper semicircular disk of radius \(5\).
Each additional equation or inequality narrows the feasible set.
Locus Problems as Translation Problems
Many locus problems can be solved systematically by translating words into algebra.
Typical phrases include:
- “exactly \(r\) units from” \(\rightarrow\) distance equality;
- “closer to \(A\) than \(B\)” \(\rightarrow\) distance inequality;
- “equidistant from” \(\rightarrow\) equal distances;
- “on a line” \(\rightarrow\) linear equation;
- “inside” \(\rightarrow\) inequality;
- “between” \(\rightarrow\) compound inequalities.
Common Error: Describing Only One Example Point
A locus is an entire set of points, not one representative point.
If the condition is
\[ x^2+y^2=9, \]then
\[ (3,0) \]is one point on the locus, but it is not the complete answer.
The full locus is the circle of radius \(3\).
Common Error: Confusing Boundary with Region
The equation
\[ x^2+y^2=16 \]describes only the circle.
The inequality
\[ x^2+y^2\le16 \]describes the circle and everything inside it.
Equality and inequality are geometrically different.
Common Error: Squaring Without Considering Equivalence
Squaring is safe when comparing nonnegative distance expressions, but in general it can introduce extraneous solutions.
For example,
\[ a=b \]implies
\[ a^2=b^2, \]but the reverse allows
\[ a=-b. \]Always use the geometric sign information available in distance problems.
Common Error: Forgetting Boundary Inclusion
The phrases
- less than;
- at most;
- greater than;
- at least
correspond to different inequality symbols.
In geometry, this determines whether the boundary belongs to the locus.
Common Error: Expanding Too Early
An equation such as
\[ (x-3)^2+(y+2)^2=25 \]immediately reveals a center and radius.
Expanding to general form may hide that structure.
Keep equations in geometrically meaningful forms when possible.
A Reliable Procedure for Locus Problems
- Define a general point, usually \((x,y)\) or \((x,y,z)\).
- Translate the geometric condition into an algebraic statement.
- Use distance, slope, midpoint, dot product, or another appropriate formula.
- Simplify the equation or inequality carefully.
- Identify the resulting geometric set.
- Check boundary inclusion when inequalities are involved.
- Verify that every derived point satisfies the original geometric condition.
Practice 1: Fixed Distance
Describe algebraically the locus of points exactly \(6\) units from
\[ (2,-3). \]Practice 2: Inside a Circle
Describe the set of points strictly inside the circle centered at the origin with radius \(4\).
Practice 3: Equidistant from Two Points
Find the locus of points equidistant from
\[ (-4,0) \]and
\[ (2,0). \]Practice 4: Half-Plane
Describe the region above and including the line
\[ y=3x-2. \]Practice 5: Distance from an Axis
Describe the locus of points exactly \(5\) units from the \(y\)-axis.
Practice 6: Strip
Describe the region between and including the vertical lines
\[ x=-2 \]and
\[ x=7. \]Practice 7: Union of Lines
Describe geometrically the locus
\[ (x-1)(y+4)=0. \]Practice 8: Sphere
Describe algebraically the locus of points in \(\mathbb{R}^3\) exactly \(3\) units from
\[ (1,-2,4). \]Practice 9: Region in Space
Describe geometrically
\[ x^2+y^2+z^2<16. \]Practice 10: Identify the Error
A student says
\[ x^2+y^2\le9 \]describes the circle of radius \(3\).
Explain the error.
Practice Answers
Practice 1
Use the distance formula:
\[ \boxed{(x-2)^2+(y+3)^2=36}. \]The locus is a circle centered at \((2,-3)\) with radius \(6\).
Practice 2
\[ \boxed{x^2+y^2<16}. \]This is the open disk of radius \(4\) centered at the origin.
Practice 3
Set squared distances equal:
\[ (x+4)^2+y^2=(x-2)^2+y^2. \]Expanding gives
\[ x^2+8x+16=x^2-4x+4. \]Therefore,
\[ 12x=-12, \]so
\[ \boxed{x=-1}. \]The locus is the perpendicular bisector of the two fixed points.
Practice 4
Above the line means \(y\) is greater, and including the line requires equality:
\[ \boxed{y\ge3x-2}. \]Practice 5
Distance from the \(y\)-axis is \(|x|\):
\[ |x|=5. \]Therefore,
\[ \boxed{x=5\quad\text{or}\quad x=-5}. \]Practice 6
\[ \boxed{-2\le x\le7}. \]Practice 7
By the zero-product principle,
\[ x-1=0 \]or
\[ y+4=0. \]Therefore the locus is
\[ \boxed{x=1\quad\text{or}\quad y=-4}. \]Practice 8
\[ \boxed{ (x-1)^2+(y+2)^2+(z-4)^2=9 }. \]This is a sphere centered at \((1,-2,4)\) with radius \(3\).
Practice 9
The inequality describes all points less than \(4\) units from the origin.
Therefore it is the
\[ \boxed{\text{interior of the sphere of radius }4}. \]Practice 10
The symbol
\[ \le \]includes every point inside the circle as well as the boundary.
Therefore the locus is the closed disk of radius \(3\), not merely the circle.
The circle alone is
\[ \boxed{x^2+y^2=9}. \]What to Remember
A locus is the set of all points satisfying a geometric condition.
Coordinate geometry translates those conditions into equations and inequalities.
- fixed distance from a point \(\rightarrow\) circle or sphere;
- equidistant from two points \(\rightarrow\) perpendicular bisector or perpendicular-bisector plane;
- fixed distance from an axis or line \(\rightarrow\) parallel loci;
- equal distance from focus and directrix \(\rightarrow\) parabola;
- equality \(\rightarrow\) boundary or exact locus;
- inequality \(\rightarrow\) region;
- simultaneous equations \(\rightarrow\) intersections;
- products and absolute values can encode unions and symmetry.
The key skill is translation:
\[ \boxed{ \text{geometry} \longrightarrow \text{algebra} \longrightarrow \text{geometric interpretation} }. \]Where We Go Next
Algebraic descriptions let us express geometric conditions precisely. The next step is to use those descriptions as tools for proving whether a geometric claim is true.
The next lesson, Coordinate Verification of Geometric Claims, develops systematic methods for checking parallelism, perpendicularity, congruence, midpoint relationships, collinearity, and other geometric statements using coordinates and algebra.