1. **State the problem:** Find the equation of the line passing through points C(10, 15) and D(-8, -21) in the form $y = mx + c$. 2. **Calculate the slope $m$:** Use the formula for slope between two points $(x_1, y_1)$ and $(x_2, y_2)$: $$m = \frac{y_2 - y_1}{x_2 - x_1}$$ Substitute the values: $$m = \frac{-21 - 15}{-8 - 10} = \frac{-36}{-18} = 2$$ 3. **Find the y-intercept $c$:** Use the equation $y = mx + c$ and substitute one point, say C(10, 15): $$15 = 2 \times 10 + c$$ $$15 = 20 + c$$ $$c = 15 - 20 = -5$$ 4. **Write the equation of the line:** $$y = 2x - 5$$ 5. **Check if point (-2, -9) lies on the line:** Substitute $x = -2$ into the equation: $$y = 2(-2) - 5 = -4 - 5 = -9$$ Since the calculated $y$ matches the point's $y$ coordinate, the point lies on the line. **Final answers:** (i) Slope $m = 2$, y-intercept $c = -5$ (ii) Equation of line: $$y = 2x - 5$$ (iii) Point (-2, -9) lies on the line.