MISSION · Geometry
Locus — Paths Defined by a Rule
Every point obeying one rule, and the shape they turn out to make
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The idea in one sentence
A locus is the set of every point satisfying some condition — and the striking thing is how often a simple condition produces a familiar shape.
The four standard ones
Nearly every locus question is built from these.
Fixed distance from a point. A circle, centre at the point, radius the distance. This is not so much a theorem as the definition of a circle restated.
Equidistant from two points \(A\) and \(B\). The perpendicular bisector of \(AB\). Every point on it is the same distance from both ends — which is exactly why the compass construction for a bisector works.
Fixed distance from a line. Two parallel lines, one either side. It is easy to give only one and lose half the answer.
Equidistant from two intersecting lines. The pair of angle bisectors. Two again, at right angles to each other.
Read those as a pattern rather than a list: a condition involving one object usually gives a curve around it, and a condition balancing two objects usually gives something down the middle.
Conditions that produce the conics
Push the idea slightly and the conic sections appear, each as a locus.
Equidistant from a point and a line gives a parabola. The point is its focus, the line its directrix, and that is the original definition — the \(y = x^2\) equation is a consequence of it.
Distances to two fixed points summing to a constant gives an ellipse. This is the gardener’s construction: a loop of string round two pegs, pulled taut by a pencil.
Distances to two fixed points differing by a constant gives a hyperbola, and it is how navigation systems located ships from the timing difference between two radio transmitters.
One family of curves, three distance conditions. Defining conics this way rather than by slicing a cone makes it far clearer why they turn up in orbits and in acoustics.
Turning a locus into an equation
The algebraic method is mechanical: call the point \((x, y)\), write the condition using the distance formula, and simplify.
Points equidistant from \((0,0)\) and \((4,0)\):
\[\sqrt{x^2+y^2} = \sqrt{(x-4)^2+y^2}\]Squaring both sides, \(x^2 = x^2 - 8x + 16\), so \(x = 2\) — the vertical line through the midpoint, which is the perpendicular bisector as promised.
Points 5 from the origin: \(\sqrt{x^2+y^2} = 5\), so \(x^2 + y^2 = 25\). The circle equation is just a locus condition with the root cleared.
This is coordinate proof applied to a definition, and it turns “describe the shape” into algebra you can simply do.
Regions, not just paths
Change the equals sign to an inequality and the locus becomes an area.
“Within 5 of the origin” is the filled disc \(x^2 + y^2 \le 25\). “Closer to \(A\) than to \(B\)” is the half-plane on \(A\)’s side of the perpendicular bisector.
Conventions worth following when you draw one: a solid boundary is included, a dashed boundary is excluded, and the region satisfying the condition gets shaded. Combining several conditions means shading only the overlap.
This is what mobile phone coverage maps, safety exclusion zones and the catchment areas of schools all are — loci with inequalities.
Common mix-ups
- Giving one line where there are two. Distance from a line, and the angle-bisector case, both have two branches.
- Stopping at one point. “Equidistant from \(A\) and \(B\)” is not the midpoint alone; the midpoint is one member of an infinite set.
- Forgetting that a locus is unbounded unless the condition bounds it. A perpendicular bisector runs forever.
- Drawing in only two dimensions when the question is spatial. Points 5 cm from a point in space form a sphere, not a circle.
- Shading the wrong side. Test one convenient point — the origin usually — and see whether it satisfies the condition.
Check yourself
What is the locus of points a fixed distance from a point, equidistant from two points, and equidistant from a point and a line?
Fixed distance from a point is a circle. Equidistant from two points is the perpendicular bisector of the segment joining them. And equidistant from a point and a line is a parabola, with that point as focus and that line as directrix.
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