What is the locus of the point X which moves relative to two fixed points P and M on a plane such that < PXM = 30o

Analyze the question

We want every point X such that the angle PXM, formed by joining X to the two fixed points P and M, is always 30°. We need the shape this traces out.

Concept

This comes from the circle theorem: angles in the same segment of a circle, standing on the same chord, are equal.

If a chord PM is fixed, every point X on the same arc (on the same side of PM) sees the chord at the same angle. So a constant angle PXM means X lies on an arc of a circle, with PM as a chord.

Solution

1. P and M are fixed, so the segment PM is fixed.
2. X moves so that ∠PXM = 30° at all times.
3. By the circle theorem, all points that subtend 30° on the chord PM lie on one arc of a circle passing through P and M.
4. On the other side of PM there is a mirror-image arc, which also gives 30°. So the full locus is two arcs, but "an arc of a circle with PM as a chord" is the best description among the options.

Checking the other options

• A: The bisector of line PM (perpendicular bisector) would give points equidistant from P and M, not a constant angle.
• C: The bisector of angle PXM is a line through X, not a path traced by X.
• D: A circle with centre X doesn't make sense, since X is the moving point.

Answer: B

Tip: Keep these loci in mind:

• Constant distance from a point → circle.
• Equidistant from two points → perpendicular bisector.
• Constant angle subtended by a fixed segment → arc of a circle through the two points.