1. **State the problem:** Find the equation of the perpendicular bisector of the line segment AB where A = (1,7) and B = (5,15). 2. **Find the midpoint M of AB:** The midpoint formula is $$M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$$ Substitute values: $$M = \left(\frac{1 + 5}{2}, \frac{7 + 15}{2}\right) = (3, 11)$$ 3. **Calculate the slope of AB:** The slope formula is $$m = \frac{y_2 - y_1}{x_2 - x_1}$$ Substitute values: $$m = \frac{15 - 7}{5 - 1} = \frac{8}{4} = 2$$ 4. **Find the slope of the perpendicular bisector:** The slope of a line perpendicular to another is the negative reciprocal of the original slope. So, $$m_\perp = -\frac{1}{m} = -\frac{1}{2}$$ 5. **Use point-slope form to find the equation:** The point-slope form is $$y - y_1 = m(x - x_1)$$ Using midpoint M(3,11) and slope $$m_\perp = -\frac{1}{2}$$: $$y - 11 = -\frac{1}{2}(x - 3)$$ 6. **Simplify to slope-intercept form $$y = mx + c$$:** $$y - 11 = -\frac{1}{2}x + \frac{3}{2}$$ Add 11 to both sides: $$y = -\frac{1}{2}x + \frac{3}{2} + 11 = -\frac{1}{2}x + \frac{3}{2} + \frac{22}{2} = -\frac{1}{2}x + \frac{25}{2}$$ **Final answer:** $$y = -\frac{1}{2}x + \frac{25}{2}$$ This is the equation of the perpendicular bisector of line AB in the form $$y = mx + c$$.