1. **State the problem:** Find the equation of the perpendicular bisector of the line segment joining points $A(1,-5)$ and $B(9,1)$ in the form $y=mx+c$. 2. **Find the midpoint of AB:** The midpoint $M$ has coordinates $$M=\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right) = \left(\frac{1+9}{2}, \frac{-5+1}{2}\right) = (5, -2).$$ 3. **Find the slope of AB:** The slope $m_{AB}$ is $$m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1 - (-5)}{9 - 1} = \frac{6}{8} = \frac{3}{4}.$$ 4. **Find the slope of the perpendicular bisector:** The perpendicular bisector's slope $m$ is the negative reciprocal of $m_{AB}$, so $$m = -\frac{1}{m_{AB}} = -\frac{1}{\frac{3}{4}} = -\frac{4}{3}.$$ 5. **Use point-slope form to find the equation:** Using point $M(5,-2)$ and slope $m=-\frac{4}{3}$, the equation is $$y - y_1 = m(x - x_1)$$ $$y - (-2) = -\frac{4}{3}(x - 5)$$ $$y + 2 = -\frac{4}{3}x + \frac{20}{3}.$$ 6. **Simplify to slope-intercept form:** Subtract 2 from both sides: $$y = -\frac{4}{3}x + \frac{20}{3} - 2.$$ Write 2 as $\frac{6}{3}$ to combine: $$y = -\frac{4}{3}x + \frac{20}{3} - \frac{6}{3} = -\frac{4}{3}x + \frac{14}{3}.$$ **Final answer:** $$\boxed{y = -\frac{4}{3}x + \frac{14}{3}}.$$