Lesson 8 of 10 · 7 min
Equations of sets of points
NCERT §11.4 Example 6, Example 8
Two radio beacons sit at A(1, 0, 2) and B(3, 2, 0). The drone must hover where it is equally far from both. Which points qualify?
The lesson in notes
In short
A condition on distances from fixed points picks out a set of points in space. Write P = (x, y, z), express each distance with the formula, and simplify: the result is an equation in x, y and z.
Points equally distant from A(3, 4, −5) and B(−2, 1, 4): set PA² = PB². The x², y² and z² terms cancel, leaving 10x + 6y − 18z − 29 = 0, a first-degree equation.
Squaring both sides of PA = PB is safe here because both sides are distances and so never negative.
Points with PA² + PB² = 2k² for A(3, 4, 5) and B(−1, 3, −7): adding the two squared distances gives 2x² + 2y² + 2z² − 4x − 14y + 4z = 2k² − 109.
The constant 109 is the sum of the squares of all six coordinates of A and B: 9 + 16 + 25 + 1 + 9 + 49 = 109.
Keep the variable point general. Write (x, y, z) throughout and only at the end check a known point, such as the midpoint of AB for the equal-distance set.