Loci and mixed problems

A locus is the set of all points that obey a condition. To find its equation, let P(x, y, z) be any such point, write the condition with the distance formula, square to remove roots, and simplify.

Equidistant from two points. Points equally far from A(1, 0, 2) and B(3, 2, 0):

• PA² = PB²: (x – 1)² + y² + (z – 2)² = (x – 3)² + (y – 2)² + z².
• The x², y² and z² cancel, leaving 4x + 4y – 4z – 8 = 0, that is x + y – z – 2 = 0.
• This is a plane (no squares left): the plane that cuts AB in half at right angles. The mid-point (2, 1, 1) satisfies it.

A fixed distance from one point. Points 5 units from the origin satisfy x² + y² + z² = 25: a sphere.

A fixed sum of distances. Points with PA + PB = 10, where A(3, 0, 0) and B(–3, 0, 0):

• Write PA = 10 – PB and square: the big terms cancel and leave 20·PB = 100 + 12x.
• Square again: 25[(x + 3)² + y² + z²] = (25 + 3x)².
• Simplify: 16x² + 25y² + 25z² = 400, an egg-shaped surface called an ellipsoid.

Mid-point and centroid work exactly as in 2D, with a third coordinate:

• Mid-point = the average of the two points. (2, –4, 6) and (4, 2, 0) have mid-point (3, –1, 3).
• Centroid of a triangle = the average of the three vertices. A(1, 2, 3), B(4, –1, 0), C(1, 5, –6): ((1 + 4 + 1)/3, (2 – 1 + 5)/3, (3 + 0 – 6)/3) = (2, 2, –1).

Going backwards: if A(2, 0, 1), B(–1, 3, 4) and the centroid is (2, 2, 2), then A + B + C = 3 × (2, 2, 2) = (6, 6, 6), so C = (5, 3, 1).

Fourth vertex of a parallelogram ABCD. The diagonals AC and BD cut each other in half, so they have the same mid-point: A + C = B + D, giving D = A + C – B.

• A(1, 0, 2), B(3, 1, 0), C(4, 3, 1): D = (1 + 4 – 3, 0 + 3 – 1, 2 + 1 – 0) = (2, 2, 3).
• Check: AC and BD both have mid-point (2.5, 1.5, 1.5).

A median joins a vertex to the mid-point of the opposite side. For A(0, 0, 0), B(4, 0, 0), C(0, 6, 0), the median from A goes to (2, 3, 0) and has length √(4 + 9) = √13.