1. **State the problem:** Find the equation of the line which contains the point $C$ and is perpendicular to the line $2x - 3y + 6 = 0$. 2. **Find point $C$:** From question 10(i), point $C$ is the intersection of the line $2x - 3y + 6 = 0$ with the x-axis. On the x-axis, $y=0$. Substitute $y=0$ into the line equation: $$2x - 3(0) + 6 = 0 \implies 2x + 6 = 0$$ $$2x = -6$$ $$x = -3$$ So, $C = (-3, 0)$. 3. **Find the slope of the given line:** Rewrite $2x - 3y + 6 = 0$ in slope-intercept form $y = mx + b$. $$2x - 3y + 6 = 0 \implies -3y = -2x - 6$$ $$y = \frac{2}{3}x + 2$$ The slope of the given line is $m = \frac{2}{3}$. 4. **Find the slope of the perpendicular line:** The slope of a line perpendicular to another with slope $m$ is the negative reciprocal: $$m_{\perp} = -\frac{1}{m} = -\frac{1}{\frac{2}{3}} = -\frac{3}{2}$$ 5. **Use point-slope form to find the equation:** The line passes through $C(-3, 0)$ with slope $m_{\perp} = -\frac{3}{2}$. $$y - y_1 = m(x - x_1)$$ $$y - 0 = -\frac{3}{2}(x - (-3))$$ $$y = -\frac{3}{2}(x + 3)$$ 6. **Simplify the equation:** $$y = -\frac{3}{2}x - \frac{9}{2}$$ 7. **Convert to standard form:** Multiply both sides by 2 to clear denominators: $$2y = -3x - 9$$ Rearranged: $$3x + 2y + 9 = 0$$ **Final answer:** The equation of the line perpendicular to $2x - 3y + 6 = 0$ passing through $C(-3, 0)$ is $$3x + 2y + 9 = 0$$