a locus
/ LOH-kus /
A locus is the set of all points that obey a stated geometric condition — the 'place' where every point satisfying a rule lives, and nowhere else. The word is Latin for 'place', and its plural is loci. Whenever you describe a figure by a property rather than by listing its points ('all points equidistant from a fixed point'), you are describing a locus.
Coordinate geometry gives the locus idea its power by translating the condition into an equation: a point (x, y) belongs to the locus exactly when its coordinates satisfy the equation, so the locus is precisely the solution set of that equation, and the curve you draw is a picture of that set. The recipe is: let a general point be (x, y), write the given condition using distance, slope, or whatever the condition mentions, then simplify to an equation in x and y. For example, 'all points 5 units from the origin' becomes sqrt(x^2 + y^2) = 5, i.e. x^2 + y^2 = 25 — a circle. 'All points equidistant from two fixed points' simplifies to the equation of their perpendicular bisector, a straight line.
This 'condition becomes equation, equation becomes curve' dictionary is the whole engine of analytic geometry: circles, the conic sections, the perpendicular bisector, and the angle bisector are all loci you can pin down algebraically. A useful caution: a locus is the complete set of qualifying points — neither too few nor too many — so when you derive its equation you should check that every solution really meets the condition and no qualifying point was lost, since squaring or clearing fractions can occasionally add or drop points.
Find the locus of points equidistant from A(-2, 0) and B(4, 0). Let P = (x, y); |PA| = |PB| means (x + 2)^2 + y^2 = (x - 4)^2 + y^2. Expand and cancel y^2: x^2 + 4x + 4 = x^2 - 8x + 16, so 12x = 12, x = 1 — the vertical line through the midpoint, exactly the perpendicular bisector of AB.
A geometric condition (equidistant) turns into an algebraic equation (x = 1), whose graph is the locus.
A locus is exactly the qualifying points — no more, no less. When you square both sides to remove a root, recheck the result, since squaring can introduce extra solutions that do not actually satisfy the original condition.