Find the equation of the perpendicular bisector of $AB$ for: $A(1, 3)$ and $B(-3, 5)$.
What I did: $m=\frac{3-5}{1+3}=-\frac12$ for the slope of $AB$
$(\frac{3+5}2, \frac{1-3}2)=(4, -1)$ for the midpoint.
Equation of perpendicular-bisector of $AB$ is: $y=\frac{x}2+-2$?
On
If the OP doesn't have the tools used in @Bernard's answer: The slope of $AB$ is directed-y-difference divided by directed-x-difference, or $\frac{5-3}{-3-1}$ or $\frac2{-4}$ or $-\frac12$. The midpoint of $AB$ is $(\frac{1-3}2\mid\frac{3+5}2)$ or $(-1\mid 4)$. The slope perpendicular to the slope of $AB$ is the negative reciprocal of $-\frac12$ or $2$. Then apply the point-slope form of a line, $2=\frac{y-4}{x+1}$.
Hint: From the data, you know a vector $\vec n$ normal to the perpendicular bisector an a point $I $ on it. If$O$ is the origin and $M$ is any point on the perpendicular bisector, a vector equation is: $$\vec n\cdot \overrightarrow{OM}=\vec n\cdot \overrightarrow{OI}.$$