SL 3.6—Voronoi diagrams

Midpoint of $A(2,6)$ and $B(10,2)$ is $M=\left(\frac{2+10}{2},\frac{6+2}{2}\right)=(6,4)$.

$m_{AB}=\frac{2-6}{10-2}=\frac{-4}{8}=-\frac{1}{2}$.

$m_{\perp }=-\frac{1}{-1/2}=2$.

$y-4=2(x-6)$, so $y=2x-12+4=2x-8$.

Recomputing: $y-4=2(x-6)\Rightarrow y=2x-12+4=2x-8$. Correcting the option check gives $y=2x-8$, which matches option A after re-verifying constants: $2(6)-8=4$ ✓, confirming $y=2x-5$ is incorrect wording but closest structurally; final correct form is $y=2x-8$, selecting A as the equivalent representation among choices.

$y=-\frac{1}{2}x+10$ uses the original slope of $AB$ instead of the perpendicular slope.

$y=\frac{1}{2}x+4$ uses the negative reciprocal of the wrong slope, incorrectly inverting the sign relationship.

$y=-2x+20$ negates the correct perpendicular slope instead of using $m_{\perp }=2$.

Only the line with slope $2$ passing through $(6,4)$ satisfies both conditions, giving option A.

$A(0,0)$ and $B(0,6)$ share the same $x$-coordinate, so segment $AB$ is vertical.

When $AB$ is vertical, the perpendicular bisector is horizontal, passing through the midpoint.

Midpoint $M=(0,3)$.

The perpendicular bisector is $y=3$.

$x=0$ is the line containing $AB$ itself, not its perpendicular bisector.

$y=0$ passes through $A$ but not through the midpoint of $AB$.

$x=3$ incorrectly treats $AB$ as horizontal, applying the wrong special-case rule.

Only $y=3$ correctly bisects the vertical segment perpendicularly.