Line segment $P Q$ has endpoints at $P(6,-2)$ and $Q(-3,10)$. Point $R(a, b)$ is a point on $P Q$ such that the distance from $P$ to $R$ is one-third the distance from $P$ to $Q$. What is $b-a$ ?
To get from $P(6,-2)$ to $Q(-3,10)$, we go $6-(-3)=9$ units to the left and $10-(-2)=12$ units up.

One-third of this difference would be 3 units to the left and 4 units up, which means that the coordinates of $R$ are $(a, b)=(6-3,-2+4)=(3,2)$.

Finally, $b-a=2-3=-1$.

ANSWER: -1