|z − A| = |z − B| Is the Perpendicular Bisector of AB

For a complex number $z = x + iy$, the modulus $|z - z_0|$ is exactly the Euclidean distance from the point $z$ to the fixed point $z_0$ in the plane. So whenever a konkur problem writes a modulus equation, translate it into a sentence about distances — the geometry then reads off instantly.

The core idea

The equation

$|z - A| = |z - B|$

says "$z$ is equidistant from the two fixed points $A$ and $B$." The set of all points equidistant from two distinct points is, by definition, the perpendicular bisector of the segment $AB$: the straight line that passes through the midpoint of $AB$ and is perpendicular to it.

Note this is a line, never a circle. A circle appears only when the two sides have different coefficients, e.g. $|z - A| = k\,|z - B|$ with $k \ne 1$ (an Apollonius circle). With coefficient $1$ on both sides you always get a line.

How to apply it (two routes)

Geometric route (fast). Identify the points $A$ and $B$. The locus is the perpendicular bisector of $\overline{AB}$:

• It passes through the midpoint $M = \tfrac{A+B}{2}$.
• Its slope is the negative reciprocal of the slope of $AB$.
• Special cases worth memorizing: if $A,B$ share the same real part (vertical segment), the bisector is a horizontal line $y = \operatorname{Im}(M)$; if they share the same imaginary part (horizontal segment), the bisector is a vertical line $x = \operatorname{Re}(M)$.

Algebraic route (safe). Put $z = x + iy$, square both sides to kill the square roots, and expand. The $x^2$ and $y^2$ terms always cancel, leaving a linear equation in $x$ and $y$ — confirming the locus is a line.

Worked mini-example

Find the locus of $|z - 2i| = |z - 4|$. Here $A = 2i = (0, 2)$ and $B = 4 = (4, 0)$. Set $z = x + iy$ and square:

$x^2 + (y-2)^2 = (x-4)^2 + y^2.$

Expanding, $x^2 + y^2 - 4y + 4 = x^2 - 8x + 16 + y^2$. Cancel $x^2$ and $y^2$:

$-4y + 4 = -8x + 16 \;\Longrightarrow\; 8x - 4y - 12 = 0 \;\Longrightarrow\; y = 2x - 3.$

Check: the midpoint $M = (2, 1)$ satisfies $1 = 2(2) - 3$, and the slope of $AB$ is $\tfrac{0-2}{4-0} = -\tfrac12$, whose negative reciprocal is $2$ — exactly the slope we found.

Exam trick: read $|z - A| = |z - B|$ as "distance to $A$ = distance to $B$" and the answer is a line — the perpendicular bisector — before doing any algebra. If $A$ and $B$ differ only in their imaginary parts the line is horizontal ($y = \text{const}$); only in real parts, vertical ($x = \text{const}$). A coefficient $\ne 1$ on one modulus would turn it into a circle, so watch for that.