1. **State the problem:** Find the equation of a straight line passing through the point $(-2,-1)$ and perpendicular to the line given by $6x - 4y = 8$. 2. **Rewrite the given line in slope-intercept form:** $$6x - 4y = 8 \implies -4y = -6x + 8 \implies y = \frac{6}{4}x - 2 = \frac{3}{2}x - 2$$ The slope of the given line is $m_1 = \frac{3}{2}$. 3. **Find the slope of the perpendicular line:** The slope of a line perpendicular to another with slope $m_1$ is $m_2 = -\frac{1}{m_1}$. So, $$m_2 = -\frac{1}{\frac{3}{2}} = -\frac{2}{3}$$ 4. **Use point-slope form to find the equation of the required line:** The point-slope form is: $$y - y_1 = m(x - x_1)$$ Using point $(-2,-1)$ and slope $m_2 = -\frac{2}{3}$: $$y - (-1) = -\frac{2}{3}(x - (-2))$$ $$y + 1 = -\frac{2}{3}(x + 2)$$ 5. **Simplify the equation:** $$y + 1 = -\frac{2}{3}x - \frac{4}{3}$$ $$y = -\frac{2}{3}x - \frac{4}{3} - 1 = -\frac{2}{3}x - \frac{4}{3} - \frac{3}{3} = -\frac{2}{3}x - \frac{7}{3}$$ **Final answer:** $$\boxed{y = -\frac{2}{3}x - \frac{7}{3}}$$