Given a set $P$ of $n$ points and a set $S$ of $m$ weighted disks in the plane, the disk coverage problem asks for a subset of disks of minimum total weight that cover all points of $P$. The problem is NP-hard. In this paper, we consider a line-constrained version in which all disks are centered on a line $L$ (while points of $P$ can be anywhere in the plane). We present an $O((m+n)\log(m+n)+κ\log m)$ time algorithm for the problem, where $κ$ is the number of pairs of disks that intersect. Alternatively, we can also solve the problem in $O(nm\log(m+n))$ time. For the unit-disk case where all disks have the same radius, the running time can be reduced to $O((n+m)\log(m+n))$. In addition, we solve in $O((m+n)\log(m+n))$ time the $L_{\infty}$ and $L_1$ cases of the problem, in which the disks are squares and diamonds, respectively. As a by-product, the 1D version of the problem where all points of $P$ are on $L$ and the disks are line segments on $L$ is also solved in $O((m+n)\log(m+n))$ time. We also show that the problem has an $Ω((m+n)\log (m+n))$ time lower bound even for the 1D case. We further demonstrate that our techniques can also be used to solve other geometric coverage problems. For example, given in the plane a set $P$ of $n$ points and a set $S$ of $n$ weighted half-planes, we solve in $O(n^4\log n)$ time the problem of finding a subset of half-planes to cover $P$ so that their total weight is minimized. This improves the previous best algorithm of $O(n^5)$ time by almost a linear factor. If all half-planes are lower ones, then our algorithm runs in $O(n^2\log n)$ time, which improves the previous best algorithm of $O(n^4)$ time by almost a quadratic factor.