1. **Problem Statement:** Find the equation of the line through the midpoint of segment AB that is perpendicular to AB, where A(-2, -3) and B(4, 9). Then use this to find the coordinates of the circumcentre of triangle ABC with C(-4, 3). 2. **Step 1: Find midpoint of AB.** The midpoint formula is $$\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$$. Midpoint M of AB: $$M = \left(\frac{-2+4}{2}, \frac{-3+9}{2}\right) = \left(\frac{2}{2}, \frac{6}{2}\right) = (1, 3)$$ 3. **Step 2: Find slope of AB.** Slope formula: $$m = \frac{y_2 - y_1}{x_2 - x_1}$$ $$m_{AB} = \frac{9 - (-3)}{4 - (-2)} = \frac{12}{6} = 2$$ 4. **Step 3: Find slope of line perpendicular to AB.** The slope of a line perpendicular to another with slope $m$ is $$m_{\perp} = -\frac{1}{m}$$. $$m_{\perp} = -\frac{1}{2}$$ 5. **Step 4: Write equation of perpendicular line through midpoint M(1,3).** Using point-slope form: $$y - y_1 = m(x - x_1)$$ $$y - 3 = -\frac{1}{2}(x - 1)$$ Simplify: $$y - 3 = -\frac{1}{2}x + \frac{1}{2}$$ $$y = -\frac{1}{2}x + \frac{1}{2} + 3 = -\frac{1}{2}x + \frac{7}{2}$$ 6. **Step 5: Find slope of BC.** Points B(4,9) and C(-4,3): $$m_{BC} = \frac{3 - 9}{-4 - 4} = \frac{-6}{-8} = \frac{3}{4}$$ 7. **Step 6: Find midpoint of BC.** $$M_{BC} = \left(\frac{4 + (-4)}{2}, \frac{9 + 3}{2}\right) = (0, 6)$$ 8. **Step 7: Find slope of perpendicular bisector of BC.** $$m_{\perp BC} = -\frac{1}{m_{BC}} = -\frac{1}{\frac{3}{4}} = -\frac{4}{3}$$ 9. **Step 8: Equation of perpendicular bisector of BC through (0,6):** $$y - 6 = -\frac{4}{3}(x - 0)$$ $$y = -\frac{4}{3}x + 6$$ 10. **Step 9: Find circumcentre by solving system of perpendicular bisectors:** Equations: $$y = -\frac{1}{2}x + \frac{7}{2}$$ $$y = -\frac{4}{3}x + 6$$ Set equal: $$-\frac{1}{2}x + \frac{7}{2} = -\frac{4}{3}x + 6$$ Multiply both sides by 6 to clear denominators: $$6 \times \left(-\frac{1}{2}x + \frac{7}{2}\right) = 6 \times \left(-\frac{4}{3}x + 6\right)$$ $$-3x + 21 = -8x + 36$$ Add $8x$ to both sides: $$-3x + 8x + 21 = 36$$ $$5x + 21 = 36$$ Subtract 21: $$5x = 15$$ Divide by 5: $$x = \cancel{\frac{5x}{5}} = \cancel{\frac{15}{5}} = 3$$ 11. **Step 10: Substitute $x=3$ into one equation to find $y$.** Using $$y = -\frac{1}{2}x + \frac{7}{2}$$: $$y = -\frac{1}{2} \times 3 + \frac{7}{2} = -\frac{3}{2} + \frac{7}{2} = \frac{4}{2} = 2$$ 12. **Final answer:** - Equation of line through midpoint of AB perpendicular to AB: $$y = -\frac{1}{2}x + \frac{7}{2}$$ - Coordinates of circumcentre of triangle ABC: $$(3, 2)$$