The orthocenter of a triangle is the concurrent intersection of the three (possibly extended) altitudes. What is the distance from the origin to the orthocenter of the triangle whose vertices are $A(-3,10)$, $B(1,2)$, and $C(9,2)$?

1. $B$ and $C$ both have $y$-coordinate $2$, so $\overline{BC}$ is horizontal and the altitude from $A$ is the vertical line $x=-3$.
2. The slope of $AC$ is $\dfrac{2-10}{9-(-3)}=-\dfrac23$, so the altitude from $B$ has slope $\dfrac32$ and equation $y-2=\dfrac32(x-1)$.
3. At $x=-3$: $y=2+\frac32(-4)=-4$, so the orthocenter is $H=(-3,-4)$. (Check: $AB$ has slope $-2$, so the altitude from $C$ is $y-2=\frac12(x-9)$, which gives $y=-4$ at $x=-3$.)
4. The distance from the origin is $\sqrt{(-3)^2+(-4)^2}=\boxed{5}$.