1. **State the problem:** We need to find the equation of a line $L_2$ that is perpendicular to the line $L_1$ given by $2y = 6x - 5$ and passes through the point $(9, -1)$. The answer should be in the form $ay + bx = c$. 2. **Rewrite the equation of $L_1$ in slope-intercept form:** $$2y = 6x - 5 \implies y = \frac{6x - 5}{2} = 3x - \frac{5}{2}$$ The slope of $L_1$ is $m_1 = 3$. 3. **Find the slope of $L_2$:** Lines perpendicular to each other have slopes that are negative reciprocals: $$m_2 = -\frac{1}{m_1} = -\frac{1}{3}$$ 4. **Use point-slope form for $L_2$ passing through $(9, -1)$:** $$y - y_1 = m_2(x - x_1)$$ $$y - (-1) = -\frac{1}{3}(x - 9)$$ $$y + 1 = -\frac{1}{3}x + 3$$ 5. **Rewrite in standard form $ay + bx = c$:** Multiply both sides by 3 to clear the fraction: $$3(y + 1) = 3\left(-\frac{1}{3}x + 3\right)$$ $$3y + 3 = -x + 9$$ Add $x$ to both sides: $$x + 3y + 3 = 9$$ Subtract 3 from both sides: $$x + 3y = 6$$ 6. **Final answer:** $$\boxed{x + 3y = 6}$$ This is the equation of the line $L_2$ perpendicular to $L_1$ and passing through $(9, -1)$ in the form $ay + bx = c$.