1. **State the problem:** We need to find the equation of the line passing through point $R(5, 2)$ and parallel to the line through points $P(-2, 1)$ and $Q(6, -1)$. The equation should be in the form $ax + by = c$ where $a$, $b$, and $c$ are integers. 2. **Find the slope of line $PQ$:** The slope formula is $m = \frac{y_2 - y_1}{x_2 - x_1}$. Calculate: $$m_{PQ} = \frac{-1 - 1}{6 - (-2)} = \frac{-2}{8} = -\frac{1}{4}$$ 3. **Slope of the line parallel to $PQ$:** Parallel lines have equal slopes, so the slope of the line through $R$ is also $m = -\frac{1}{4}$. 4. **Use point-slope form to find the equation:** $$y - y_1 = m(x - x_1)$$ Substitute $m = -\frac{1}{4}$ and point $R(5, 2)$: $$y - 2 = -\frac{1}{4}(x - 5)$$ 5. **Simplify the equation:** $$y - 2 = -\frac{1}{4}x + \frac{5}{4}$$ $$y = -\frac{1}{4}x + \frac{5}{4} + 2$$ $$y = -\frac{1}{4}x + \frac{5}{4} + \frac{8}{4} = -\frac{1}{4}x + \frac{13}{4}$$ 6. **Convert to standard form $ax + by = c$ with integers:** Multiply both sides by 4 to clear denominators: $$4y = -x + 13$$ Rewrite: $$x + 4y = 13$$ 7. **Final answer:** The equation of the line passing through $R$ and parallel to $PQ$ is: $$\boxed{x + 4y = 13}$$