Mathematics.

coordinate geometry

Loci in Analytic Geometry

Analytic Geometry25 minDifficulty4 out of 10

You should know: distance and midpoint formulas

Overview

A locus (plural: loci) is the set of all points satisfying a given geometric condition — the word comes from the Latin for 'place.' Analytic geometry turns a locus problem into algebra: state the condition as an equation relating the coordinates (x, y) of an arbitrary point on the locus, then simplify. This single technique — translate a geometric description into a distance or slope condition, then reduce algebraically — is exactly how the standard equations of circles, parabolas, ellipses, and hyperbolas were originally derived, and it remains the general method for finding the equation of any new geometric locus.

Intuition

Rather than trying to visualize the whole shape at once, imagine a single generic point P=(x,y) that is required to obey some rule — 'stay 5 units from the origin,' 'stay equidistant from two fixed points,' 'stay twice as far from one point as from a line.' Write that rule using the distance formula, and the resulting equation in x and y is automatically satisfied by every point obeying the rule and no others. Simplifying that equation (often by squaring both sides to clear a square root, then completing the square) reveals the shape hiding inside the condition — a circle, a perpendicular bisector, a parabola, or something more exotic.

Formal Definition

Definition

To find the locus of points satisfying a condition, let P = (x, y) be an arbitrary point satisfying that condition, express the condition using the distance formula (or other coordinate relations), and simplify to obtain an equation in x and y:

d(P,A)=(x−a1)2+(y−a2)2d(P, A) = \sqrt{(x-a_1)^2 + (y-a_2)^2}
Distance from P=(x,y) to a fixed point A=(a₁,a₂)
Locus={(x,y):condition(x,y) holds}\text{Locus} = \{(x,y) : \text{condition}(x,y) \text{ holds}\}
General definition of a locus

Derivation

As a concrete derivation, find the locus of points equidistant from two fixed points A=(a₁,a₂) and B=(b₁,b₂) — this is the classic perpendicular bisector.

(x−a1)2+(y−a2)2=(x−b1)2+(y−b2)2\sqrt{(x-a_1)^2+(y-a_2)^2} = \sqrt{(x-b_1)^2+(y-b_2)^2}

Equal distance to A and B

(x−a1)2+(y−a2)2=(x−b1)2+(y−b2)2(x-a_1)^2+(y-a_2)^2 = (x-b_1)^2+(y-b_2)^2

Square both sides to remove the radicals

−2a1x+a12−2a2y+a22=−2b1x+b12−2b2y+b22-2a_1x+a_1^2-2a_2y+a_2^2 = -2b_1x+b_1^2-2b_2y+b_2^2

Expand; the x² and y² terms cancel from both sides

2(b1−a1)x+2(b2−a2)y=b12+b22−a12−a222(b_1-a_1)x + 2(b_2-a_2)y = b_1^2+b_2^2-a_1^2-a_2^2

Collect terms — a single linear equation, confirming the locus is a straight line (the perpendicular bisector of AB)

Properties

Equidistant from two points

The locus is the perpendicular bisector of the segment joining the two points — always a straight line\text{The locus is the perpendicular bisector of the segment joining the two points — always a straight line}

Equidistant from a point and a line

The locus is a parabola, with the point as focus and the line as directrix\text{The locus is a parabola, with the point as focus and the line as directrix}

Fixed distance from a point

The locus is a circle centered at that point\text{The locus is a circle centered at that point}

Constant sum of distances to two foci

The locus is an ellipse (when the constant sum exceeds the distance between the foci)\text{The locus is an ellipse (when the constant sum exceeds the distance between the foci)}

Applications

Triangulation and trilateration systems (GPS, radio navigation) locate a position as the locus of points at specified distances from known reference points.

Worked Examples

  1. Set distances equal and square both sides.

    (x−2)2+(y−3)2=(x−6)2+(y+1)2(x-2)^2+(y-3)^2 = (x-6)^2+(y+1)^2

  2. Expand both sides.

    x2−4x+4+y2−6y+9=x2−12x+36+y2+2y+1x^2-4x+4+y^2-6y+9 = x^2-12x+36+y^2+2y+1

  3. Cancel x² and y², then collect like terms.

    −4x−6y+13=−12x+2y+37-4x-6y+13 = -12x+2y+37

  4. Move everything to one side and simplify.

    8x−8y−24=0  ⟹  x−y−3=08x - 8y - 24 = 0 \implies x - y - 3 = 0

Answer: The locus is the line x − y − 3 = 0 (equivalently y = x − 3), the perpendicular bisector of segment AB.

Practice Problems

Difficulty 4/10

Find the locus of points equidistant from (0, 0) and (0, 8).

Difficulty 5/10

Find the locus of points at a fixed distance 5 from the point (3, -2).

Difficulty 6/10

Two radio towers are located at A = (-4, 0) and B = (4, 0) (in km). A ship records that it is always 6 km closer to tower A than to tower B (that is, distance to B minus distance to A equals 6). This locus is a hyperbola branch with foci A, B. Find 'a' (half the constant difference) and c (half the distance between foci), then b².

Common Mistakes

Common Mistake

Forgetting to square both sides correctly when eliminating a square root, e.g. dropping a cross term from (y-8)².

Expand squared binomials fully: (y-8)² = y² - 16y + 64, not y² + 64. Skipping the middle term is a common algebra slip that changes the final locus entirely.

Common Mistake

Assuming every locus problem produces a conic section.

Many loci are not conics at all — equidistance from two points gives a straight line, and more complex conditions can give cubic curves, or other shapes entirely. Conics arise specifically from distance/directrix-type conditions.

Quiz

The locus of points equidistant from two fixed points is always:
The locus of points equidistant from a fixed point (focus) and a fixed line (directrix) is:
The standard method for finding a locus in analytic geometry begins by:

Summary

  • A locus is the set of all points satisfying a stated geometric condition, found by writing that condition algebraically for a generic point P=(x,y).
  • Equidistance from two fixed points gives a straight line (the perpendicular bisector); equidistance from a point and a line gives a parabola.
  • Fixed distance from a point gives a circle; the standard conics (ellipse, hyperbola) arise from sum/difference-of-distances conditions to two foci.
  • The general method — state the condition with distances, square to clear radicals, simplify — is how every standard conic equation was originally derived.

References

  1. WebsiteWikipedia — Locus (mathematics)