Finding the Equation of a Locus

Two conditions come up repeatedly in Business Mathematics problems on locus: a point equidistant from two fixed points, and a point equidistant from a fixed point and a fixed line.

Equidistant from two fixed points $A$ and $B$. If $PA = PB$, squaring both sides of the distance-formula equation and simplifying always collapses the $x^2$ and $y^2$ terms, leaving a first-degree equation in $x, y$ — that is, the locus is always a straight line. That line is, geometrically, the perpendicular bisector of segment $AB$, so it must pass through the midpoint of $AB$ and be perpendicular to it; this gives a fast way to check any answer. …