323. The Locus Explorer · Where do points go when they follow the rules?

Try to break it

Drag P along its dashed line — PA and PB always stay equal, so the dashed line is the perpendicular bisector of AB. Drag Q along its dashed line — Q stays equidistant from the two crossing rays, so Q's dashed line is the angle bisector. Step a point off its line (in your head) and the equality breaks instantly.

How you build it

Construct both loci: the perpendicular bisector of two points and the angle bisector of two lines.

• Point tool: mark A on the upper-left — the first fixed point.
• Point tool: mark B below A — the second fixed point.
• Segment tool: join A to B.
• Arc tool: centre A, open it to more than half of AB, and swing an arc that reaches across to the right of AB.
• Arc tool: keep the same radius, centre B, and swing an arc. It cuts the first arc at two points.
• Line tool: click the two points where the arcs cross. This line is the perpendicular bisector of AB — every point on it is equidistant from A and B.
• Point tool: mark O on the right side — the vertex where two lines will cross.
• Line tool: click O and a point up-and-right to draw the first line through O.
• Line tool: click O and a point up-and-left to draw the second line. The two lines now cross at O.
• Arc tool: centre O, swing an arc that crosses BOTH lines — it meets them at equal distances from O.
• Point tool: mark X where the arc crosses the first line (OX is the arc radius).
• Point tool: mark Y where the arc crosses the second line. OY equals OX, so X and Y are the same distance from O.
• Bisector tool: click X then Y. Because OX = OY, the perpendicular bisector of XY passes through O and splits the angle in two — it is the locus of points equidistant from the two lines.

The proof, step by step

Prove that P traces the perpendicular bisector of AB and Q traces the angle bisector.

• Let P be any point on the constructed line. Join PA and PB.
• The arcs were drawn with equal radii, so the distances from A and B to the arc crossings are equal.
• The constructed line bisects AB at M, so AM = BM. PM is common to both triangles.
• By SSS congruence, triangle PAM ≅ triangle PBM. Therefore PA = PB.
• Thus, every point on this line is equidistant from A and B. The locus is proven.

Worked example

A point moves in a plane such that it is always equidistant from two parallel lines. The locus of the point is:

The locus of a point equidistant from two parallel lines is a line parallel to both and exactly halfway between them. This is standard Locus Rule #4.

• A circle
• A line parallel to the given lines and midway between them — correct
• A pair of angle bisectors
• A perpendicular bisector of the segment joining the lines