1. The problem asks us to find the equation of a line parallel to the line $y = -\frac{1}{3}x + 1$ that passes through the point $(1, -2)$. 2. Recall that parallel lines have the same slope. The slope of the given line is $m = -\frac{1}{3}$. 3. Use the point-slope form of a line equation: $$y - y_1 = m(x - x_1)$$ where $(x_1, y_1)$ is the point the line passes through and $m$ is the slope. 4. Substitute $m = -\frac{1}{3}$ and the point $(1, -2)$: $$y - (-2) = -\frac{1}{3}(x - 1)$$ 5. Simplify the equation: $$y + 2 = -\frac{1}{3}x + \frac{1}{3}$$ 6. Subtract 2 from both sides to solve for $y$: $$y = -\frac{1}{3}x + \frac{1}{3} - 2$$ 7. Convert $2$ to thirds to combine: $$2 = \frac{6}{3}$$ 8. So, $$y = -\frac{1}{3}x + \frac{1}{3} - \frac{6}{3} = -\frac{1}{3}x - \frac{5}{3}$$ **Final answer:** $$y = -\frac{1}{3}x - \frac{5}{3}$$