1. **State the problem:** We have line P with equation $y = -3x + 5$ and line Q is perpendicular to line P. They intersect at point $(-2, 1)$. We need to find the equation of line Q. 2. **Recall the slope rule for perpendicular lines:** If two lines are perpendicular, the slope of one line is the negative reciprocal of the other. The slope of line P is $m_P = -3$. 3. **Find the slope of line Q:** The slope of line Q, $m_Q$, is the negative reciprocal of $-3$: $$m_Q = -\frac{1}{m_P} = -\frac{1}{-3} = \frac{1}{3}$$ 4. **Use point-slope form to find the equation of line Q:** The point-slope form is: $$y - y_1 = m(x - x_1)$$ where $(x_1, y_1)$ is the point of intersection $(-2, 1)$ and $m = \frac{1}{3}$. 5. **Substitute values:** $$y - 1 = \frac{1}{3}(x - (-2)) = \frac{1}{3}(x + 2)$$ 6. **Simplify:** $$y - 1 = \frac{1}{3}x + \frac{2}{3}$$ 7. **Add 1 to both sides:** $$y = \frac{1}{3}x + \frac{2}{3} + 1$$ 8. **Convert 1 to fraction:** $$1 = \frac{3}{3}$$ 9. **Add fractions:** $$y = \frac{1}{3}x + \frac{2}{3} + \frac{3}{3} = \frac{1}{3}x + \frac{5}{3}$$ **Final answer:** $$\boxed{y = \frac{1}{3}x + \frac{5}{3}}$$