FG2, HG2 - Loci and regions

Fixed distance

A sprinkler reaches exactly $3\,\mathrm{m}$ from its centre in every direction. The outer edge is a circle, but the watered area includes positions closer to the centre too. The wording changes the answer from a line to a region.

All points exactly distance $r$ from a fixed point $A$ form a circle with centre $A$ and radius $r$. Construct it by setting the compass to $r$. “Less than $r$” means inside the circle, excluding its circumference; “at most $r$” includes the circumference. “At least $r$” means on or outside it.

[DIAGRAM: asset_name: Fixed-distance loci; asset_slug: fixed-distance-loci; caption: Distance from a point gives a circle; distance from a line gives parallels; a finite segment needs rounded ends.; description: Three panels show a circle of radius r about A, two parallel lines at perpendicular distance d on opposite sides of line l, and the locus exactly distance d from a finite segment AB, consisting of two parallel segments joined by semicircles centred at its endpoints.]

The distance from a point to a straight line is measured perpendicularly. It is the shortest route to that line: a circle expanding from the point first touches the line directly below it, along the perpendicular. A sloping route reaches another point on the line and is longer. You do not need to calculate that sloping length to identify the shortest route.

For a whole straight line, points exactly distance $d$ away lie on two parallel lines, one on each side. Construct a perpendicular, mark $d$ on both sides, then construct parallels through the marks by making lines perpendicular to that perpendicular. For distance from a finite segment, include semicircular ends: beyond an endpoint, the nearest point of the segment is that endpoint. Two infinite parallels alone would wrongly include points far beyond the segment.

Change the fixed object from a point to a line; the distance is now perpendicular separation.

A region within distance $d$ of a whole line is the strip between those parallels, including its boundaries if the wording is “at most”.

Distinguish two closely related locus rules.

Equidistant points

Equal distances from two points produce a different locus from a fixed distance from one point.

For fixed points $A$ and $B$, their perpendicular bisector is the locus equally distant from both. Construct equal arcs from $A$ and $B$, using a radius greater than half $AB$, then join the two intersections. The whole resulting line is the locus, extending beyond the short piece on your drawing.

Why is it complete? If $M$ is the midpoint and $P$ lies on the perpendicular bisector, right triangles $PMA$ and $PMB$ have equal legs $PM$ and $MA=MB$, so $PA=PB$. Conversely, if $PA=PB$, the triangles match by their three sides. Their angles at $M$ are equal and fill a straight angle, so each is $90^\circ$. Thus any such $P$ lies on the perpendicular bisector, including the midpoint itself.

[DIAGRAM: asset_name: Equal-distance bisector loci; asset_slug: equidistant-bisectors; caption: The perpendicular bisector compares distances to points; an angle bisector compares perpendicular distances to lines.; description: Left panel shows points A and B, their midpoint M, a perpendicular bisector and a point P on it with equal segments PA and PB. Right panel shows two intersecting lines forming a chosen angle, its internal angle bisector, and a point Q with equal perpendicular distances to the two lines.]

The locus nearer to $A$ than $B$ is the half-plane on $A$'s side of this bisector; the bisector itself is excluded because distances there are equal. Check the correct side by testing $A$: its distance from itself is zero.