Perpendicular bisector of a segment in coordinates

We revisit with Cartesian coordinates a result we already know from synthetic geometry: the perpendicular bisector of a segment. We find it with two different methods, which naturally lead to the same equation.

Example — Method 1: definition as a locus

Equation of the perpendicular bisector of the segment $AB$ with $A(4;5)$ and $B(1;-6)$.

The generic point $P(x_P; y_P)$ of the perpendicular bisector is equidistant from $A$ and $B$: $\sqrt{(x_P-4)^2+(y_P-5)^2} = \sqrt{(x_P-1)^2+(y_P+6)^2}$ Squaring: $x_P^2+16-8x_P+y_P^2+25-10y_P = x_P^2+1-2x_P+y_P^2+36+12y_P$ Simplifying ($x_P^2$ and $y_P^2$ cancel): $-6x_P - 22y_P = -4 \implies 3x+11y = 2 \implies \boxed{y = -\frac{3}{11}x + \frac{2}{11}}$

Example — Method 2: midpoint and perpendicularity

Midpoint: $M=\bigl(\frac{5}{2},\,-\frac{1}{2}\bigr)$.

Slope of $AB$: $m_{AB}=\dfrac{-6-5}{1-4}=\dfrac{-11}{-3}=\dfrac{11}{3}$.

Perpendicular slope: $m=-\dfrac{3}{11}$.

Line through $M$ with slope $-\frac{3}{11}$: $y+\frac{1}{2} = -\frac{3}{11}\Bigl(x-\frac{5}{2}\Bigr) \implies y = -\frac{3}{11}x+\frac{2}{11}$ Same result as Method 1. ✓

Links

Topics: Euclidean circle
Concepts: Perpendicular bisector
Skills: Analytic geometry · Using formulae