$P(3,1)$ and $R(-7,5)$ are vertices of a rhombus $PQRS$. What is the equation of diagonal $QS$?

In a rhombus, the diagonals perpendicular bisect each other. Thus, the diagonal $QS$ is the perpendicular bisector of the diagonal $PR$.Step 1: Find the midpoint of $PR$, which lies on $QS$:
$$\text{Midpoint }M=\left(\frac{3+(-7)}{2},\frac{1+5}{2}\right)=(-2,3)$$
Step 2: Find the slope of $PR$ ($m_1$):
$$m_1=\frac{5-1}{-7-3}=\frac{4}{-10}=-\frac{2}{5}$$
Step 3: Find the slope of $QS$ ($m_2$). Since $QS\perp PR$:
$$m_2=-\frac{1}{m_1}=\frac{5}{2}$$
Step 4: Find the equation of line $QS$ passing through $(-2,3)$ with slope $\frac{5}{2}$:
$$y-3=\frac{5}{2}(x-(-2))$$
$$2(y-3)=5(x+2)$$
$$2y-6=5x+10$$
$$5x-2y=-16$$