Loci and Distance Problems

From a condition to a locus

A locus is the set of all points that meet a condition. Imagine placing a pencil point exactly 3 cm from a fixed point $P$. It can move around $P$ without changing that distance: its path is a circle with centre $P$ and radius 3 cm. Set compasses to 3 cm, place the point at $P$ and draw the circle. Points less than 3 cm from $P$ lie inside it; points at most 3 cm away include the circle as well.

Distance from a line or segment

Distance from a point to a line is measured along a perpendicular: a segment meeting the line at $90^\circ$. This is the shortest route to the line. Any sloping route to another point on the line forms the longer side of a right-angled triangle. To locate points exactly 2 cm from a whole straight line, construct perpendiculars at two places, mark 2 cm on either side and draw the two parallel lines through those marks.

A finite line segment is different. Points 2 cm from its middle lie on parallel lines, but the locus curves around each endpoint. Those curved ends are semicircles of radius 2 cm centred on the endpoints. Together, the straight parts and curved ends form the boundary of all points within 2 cm of the segment. If the closest point on the segment is an endpoint, the shortest distance to the segment is to that endpoint, not to a point beyond it on the extended line.

Combining conditions with constructions

The standard ruler-and-compass constructions help you find other loci. Points equidistant from two fixed points lie on the perpendicular bisector of the segment joining them. Within a given angle, points equidistant from its two sides lie on its angle bisector; distance to each side is measured perpendicularly. Construct the relevant lines or circles first, then look for their intersections or overlapping regions.

For example, suppose a point must be exactly 4 cm from $A$ and equidistant from $A$ and $B$. Draw a circle of radius 4 cm centred on $A$ and construct the perpendicular bisector of $AB$. Their intersections are the possible positions; there may be two, one or none, depending on the measurements. If a question instead asks for points within 4 cm of $A$ and 2 cm of a segment, shade only the overlap of those two regions. The word includes the boundary, whereas does not.

You can use the same idea to construct a given figure. To make a triangle when all three side lengths are given, draw one side accurately with a ruler. From each endpoint draw a compass arc whose radius is the required length of the side from that endpoint. An intersection of the arcs gives the third vertex; join it to the endpoints. The arcs show why that vertex has both required distances. Depending on the given lengths, there may be an intersection on either side of the first side, or no possible triangle.