Equidistant points and other loci

What you will be able to do

• find the locus of the points that are equally far from two points;
• find the locus when the sum or the ratio of two distances is fixed;
• find the point that is equally far from four given points.

A beacon equally far from two towers

A beacon is to be placed so that a ship hears it equally well from tower $A$ and tower $B$. The beacon must be at the same distance from both towers. Which points work?

In a plane the answer is a straight line, the perpendicular bisector of $AB$. In space it is a plane. The method is the same for every locus: write the condition with the distance formula, square it, and simplify.

Equal distances: a plane

Let $A(1, 2, 3)$ and $B(3, -2, 1)$. A point $P(x, y, z)$ has $PA = PB$ when $PA^2 = PB^2$:

Expand both sides. The terms $x^2$, $y^2$ and $z^2$ are on both sides, so they cancel:

That simplifies to $4x - 8y - 4z = 0$, which is $x - 2y - z = 0$. All coefficients are linear, so the locus is a plane. It passes through the midpoint $(2, 0, 2)$ of $AB$, because $2 - 0 - 2 = 0$.

In general, for $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$ the plane is

The coefficients are the numbers $2(x_2 - x_1)$ and so on, which are the differences of the coordinates of $B$ and $A$. Lines and planes get a full treatment in the later chapter on three-dimensional geometry. Here we only need the equation.

Other conditions on two distances

• A fixed difference of squares: $PA^2 - PB^2 = k$. The squares cancel again, so the locus is a plane. For $k = 0$ it is the plane above.
• A fixed sum of squares: $PA^2 + PB^2 = k$. Now the squares add up, and the locus is a sphere. Its centre is the midpoint of $AB$. For $A(-2, 0, 0)$ and $B(2, 0, 0)$ we get $2(x^2 + y^2 + z^2) + 8 = k$. With $k = 40$ that is the sphere $x^2 + y^2 + z^2 = 16$.
• A fixed ratio of distances: $PA = \lambda PB$ with $\lambda \ne 1$. Squaring gives $PA^2 = \lambda^2 PB^2$, a sphere. For $A(0, 0, 0)$, $B(3, 0, 0)$ and $\lambda = 2$ we get $x^2 + y^2 + z^2 = 4\left((x - 3)^2 + y^2 + z^2\right)$. That simplifies to $x^2 + y^2 + z^2 - 8x + 12 = 0$, a sphere with centre $(4, 0, 0)$ and radius $2$. Check: $(2, 0, 0)$ has $PA = 2$ and $PB = 1$.

A point equally far from four points

If $A$, $B$, $C$, $D$ are not in one plane, there is exactly one point $P$ with $PA = PB = PC = PD$. It is the centre of the sphere through all four. Write three equations, $PA^2 = PB^2$, $PA^2 = PC^2$ and $PA^2 = PD^2$. Each is linear in $x$, $y$, $z$, so you have three linear equations to solve.

Example. Find the point equally far from $O(0, 0, 0)$, $A(2, 0, 0)$, $B(0, 4, 0)$ and $C(0, 0, 6)$.

1. From $O$ and $A$. $x^2 + y^2 + z^2 = (x - 2)^2 + y^2 + z^2$ gives $4x = 4$, so $x = 1$.
2. From $O$ and $B$. In the same way $8y = 16$, so $y = 2$.
3. From $O$ and $C$. $12z = 36$, so $z = 3$.
4. The point and the radius. $P = (1, 2, 3)$ and $PO^2 = 1 + 4 + 9 = 14$.

With three points the equal-distance points lie on a line, and with two points on a plane.

Practise with new numbers

Try it in Python

The program finds the plane of the points equidistant from $A$ and $B$, and tests four points against the equation and against the two distances.

Common mistakes

• Squaring the distances wrongly. Square both sides of $PA = PB$, so you work with $PA^2$ and $PB^2$ and no roots.
• Forgetting that the squares cancel. For $PA = PB$ and $PA^2 - PB^2 = k$ the locus is a plane. Squares stay for a sum or for a ratio.
• Expecting a line. Equal distances from two points in space make a plane, not a line.
• Losing a term in the expansion. Write each bracket out in full before you subtract.

Remember these

• $PA = PB$ gives a plane through the midpoint of $AB$, with the linear equation above.
• $PA^2 + PB^2 = k$ and $PA = \lambda PB$ ($\lambda \ne 1$) give spheres.
• Four points not in one plane have one point equally far from all: solve three linear equations.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Find the locus of the points equally far from $O(0, 0, 0)$ and $B(2, 4, -4)$.

2. Find the locus of $P$ with $PA^2 + PB^2 = 40$ for $A(-2, 0, 0)$ and $B(2, 0, 0)$.

3. Find the locus of $P$ with $PA = 2PB$ for $A(0, 0, 0)$ and $B(3, 0, 0)$.

4. Find the point equally far from $O(0, 0, 0)$, $A(2, 0, 0)$, $B(0, 4, 0)$ and $C(0, 0, 6)$. What is the common distance?

5. Does the plane $x - 2y - z = 0$ pass through the midpoint of $A(1, 2, 3)$ and $B(3, -2, 1)$?

References

• NCERT, Mathematics Textbook for Class XI, the chapter Introduction to Three Dimensional Geometry. The loci of this lesson are an extension for entrance exams.
• OpenStax, Calculus Volume 3, Chapter 2, section 2.2 on vectors in three dimensions.