1. **State the problem:** Find the equation of the line $L_2$ which 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:** $$2y = 6x - 5 \implies y = \frac{6}{2}x - \frac{5}{2} = 3x - \frac{5}{2}$$ 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. **Simplify and rearrange to the form $ay + bx = c$:** $$y + 1 = -\frac{1}{3}x + 3$$ Multiply both sides by 3 to clear the fraction: $$3y + 3 = -x + 9$$ Rearranged: $$3y + x = 9 - 3$$ $$3y + x = 6$$ 6. **Final equation of $L_2$:** $$3y + x = 6$$ --- 1. **State the problem:** Find the equation of the line $L$ with gradient 5 passing through the point $(0, -3)$. 2. **Use point-slope form:** $$y - y_1 = m(x - x_1)$$ $$y - (-3) = 5(x - 0)$$ $$y + 3 = 5x$$ 3. **Simplify to slope-intercept form:** $$y = 5x - 3$$ 4. **Final equation of $L$:** $$y = 5x - 3$$