1. **State the problem:** Find the equations of the altitudes of the triangle with vertices A(0,4), B(4,6), and C(-2,-2). 2. **Recall the definition:** An altitude of a triangle is a perpendicular line from a vertex to the opposite side. 3. **Step 1: Find the slope of each side.** - Slope of BC: $$m_{BC} = \frac{6 - (-2)}{4 - (-2)} = \frac{8}{6} = \frac{4}{3}$$ - Slope of AC: $$m_{AC} = \frac{6 - 4}{4 - 0} = \frac{2}{4} = \frac{1}{2}$$ - Slope of AB: $$m_{AB} = \frac{4 - (-2)}{0 - (-2)} = \frac{6}{2} = 3$$ 4. **Step 2: Find slopes of altitudes (perpendicular to sides).** - Altitude from A is perpendicular to BC, so slope $$m_{A} = -\frac{1}{m_{BC}} = -\frac{3}{4}$$ - Altitude from B is perpendicular to AC, so slope $$m_{B} = -\frac{1}{m_{AC}} = -2$$ - Altitude from C is perpendicular to AB, so slope $$m_{C} = -\frac{1}{m_{AB}} = -\frac{1}{3}$$ 5. **Step 3: Write equations of altitudes using point-slope form $$y - y_1 = m(x - x_1)$$** - Altitude from A(0,4): $$y - 4 = -\frac{3}{4}(x - 0) \Rightarrow y = -\frac{3}{4}x + 4$$ - Altitude from B(4,6): $$y - 6 = -2(x - 4) \Rightarrow y - 6 = -2x + 8 \Rightarrow y = -2x + 14$$ - Altitude from C(-2,-2): $$y + 2 = -\frac{1}{3}(x + 2) \Rightarrow y + 2 = -\frac{1}{3}x - \frac{2}{3} \Rightarrow y = -\frac{1}{3}x - \frac{8}{3}$$ 6. **Final answer:** - Altitude from A: $$y = -\frac{3}{4}x + 4$$ - Altitude from B: $$y = -2x + 14$$ - Altitude from C: $$y = -\frac{1}{3}x - \frac{8}{3}$$