Equation for Perpendicular Bisector of Two Points in Complex Plane/Standard Form

Theorem

Let $z_1, z_2 \in \C$ be complex numbers.

Let $L$ be the perpendicular bisector of the straight line through $z_1$ and $z_2$ in the complex plane.

$L$ can be expressed by the equation:

$\map \Re {z_2 - z_1} x + \map \Im {z_2 - z_1} y = \dfrac {\cmod {z_2}^2 - \cmod {z_1}^2} 2$

Let $z_1$ and $z_2$ be represented by the points $A = \tuple {x_1, y_1}$ and $B = \tuple {x_2, y_2}$ respectively in the complex plane.

$\ds y - \frac {y_1 + y_2} 2$

$=$

$\ds \frac {x_1 - x_2} {y_2 - y_1} \paren {x - \dfrac {x_1 + x_2} 2}$

$\ds \leadsto \ \ $

$\ds \paren {y_2 - y_1} y - \frac {\paren {y_2 - y_1} \paren {y_2 + y_1} } 2$

$=$

$\ds - \paren {x_2 - x_1} x + \frac {\paren {x_2 - x_1} \paren {x_2 + x_1} } 2$

$\ds \leadsto \ \ $

$\ds \paren {x_2 - x_1} x + \paren {y_2 - y_1} y$

$=$

$\ds \frac {x_2^2 - x_1^2 + y_2^2 - y_1^2} 2$

$\ds \leadsto \ \ $

$\ds \map \Re {z_2 - z_1} x + \map \Im {z_2 - z_1} y$

$=$

$\ds \frac {\cmod {z_2}^2 - \cmod {z_1}^2} 2$

Definition of Real Part, Definition of Imaginary Part, Definition of Complex Modulus

$\blacksquare$