1. **State the problem:** We need to find the values of $P$ and $Q$ for the line equation $Px + 2y = Q$ that passes through the point $(-3,5)$ and is perpendicular to the line $2x - y = 4$. 2. **Find the slope of the given line:** Rewrite $2x - y = 4$ in slope-intercept form $y = mx + b$. $$2x - y = 4 \implies y = 2x - 4$$ So, the slope of this line is $m = 2$. 3. **Find the slope of the perpendicular line:** The slope of a line perpendicular to another with slope $m$ is $m_{\perp} = -\frac{1}{m}$. $$m_{\perp} = -\frac{1}{2}$$ 4. **Express the line $Px + 2y = Q$ in slope-intercept form:** $$2y = -Px + Q \implies y = -\frac{P}{2}x + \frac{Q}{2}$$ So, the slope of this line is $-\frac{P}{2}$. 5. **Set the slope equal to the perpendicular slope:** $$-\frac{P}{2} = -\frac{1}{2} \implies P = 1$$ 6. **Use the point $(-3,5)$ to find $Q$:** Substitute $x = -3$, $y = 5$, and $P = 1$ into the line equation: $$1 \cdot (-3) + 2 \cdot 5 = Q \implies -3 + 10 = Q \implies Q = 7$$ **Final answer:** $$P = 1, \quad Q = 7$$