Pardon the sketchy details to follow here: there are many details that I'm burying under the rug for sake of brevity (and giving some sort of answer!). I'd recommend the first chapter of Hovey's book to get an idea for how these arguments go.
A more geometric example of this kind of argument is the construction of homotopy colimits, which are a "derived" version of colimit. Lets look at diagrams of the form $B \gets A \to C$.
Consider the two diagrams $D^n \gets S^{n-1} \to D^n$. This has colimit $S^n$, seen as gluing two disks along their boundary spheres.
Here is a weakly homotopic diagram: $* \gets S^{n-1} \to *$. This has a map into our first diagram which is object-wise a homotopy equivalence. But the colimit of this diagram is $*$, which is homotopically different from $S^n$. This means that colimit, seen as a functor from this three-space diagram category to spaces, is not a homotopy-invariant functor.
Geometrically, this can be a bit undesirable. The discrepancy arises from the fact that the maps $S^{n-1} \to *$ are not cofibrations; we're not really gluing together two points along their "common $S^{n-1}$" like we did for the pair of maps $S^{n-1} \to D^n$. Geometrically, we want the outcome to be the more interesting $S^n$.
So, we make a derived version, called homotopy colimit. It's a version of colimit that is homotopy invariant in the diagram category, but also agrees with colimit if the diagram is made up of these "nice inclusions" (cofibrations) (Again, we're sticking with these 3 space diagrams for now).
There are two ways to think about homotopy colimit that I've bumped into. One is an explicit construction: the homotopy colimit of $B \gets A \to C$ is a quotient of the coproduct $B \coprod (A \times I) \coprod C$ where we glue in $A \times 0$ to $B$ and $A \times 1$ to $C$.
If we check that with a diagram $* \gets A \to *$, we find that this gives the suspension of $A$. So in our example of $* \gets S^{n-1} \to *$, we do indeed compute $S^n$ as the homotopy colimit.
Another way to think about things is that we really wanted $D^n \gets S^{n-1} \to D^n$ to be the "correct" diagram in which to take this colimit. All the maps in this diagram are "nice" inclusions. In general, I can replace the diagram $B \gets A \to C$ with a homotopy equivalent diagram B' \gets A' \to C' where:
1) A' is a CW-complex (A "cofibrant" space)
2) A' \to B' is a CW-inclusion
3) A' \to C' is a CW-inclusion.
Then, I take homotopy colimit. So, for example, $* \gets S^{n-1} \to *$ can be replaced with $D^n \gets S^{n-1} \to D^n$ in this way.
It turns out that if I replace $B \gets A \to C$ with B' \gets A' \to C', I will not only get a colimit that does not depend on the choices I made, but also one that is homotopy invariant in the category of diagrams. A homotopy equivalent diagram $E \gets D \to F$ will produce a homotopy equivalent homotopy colimit. (Compare this to projective resolutions, where no matter which resolution I choose, I get the same derived functor value).
The replacement B' \gets A' \to C' is itself a "cofibrant" object in a model category structure on this diagram category. The model structure is constructed in such a way that colimit and the diagonal functor $A \mapsto (A \gets A \to A)$ form a Quillen adjunction -- which implies, among other things, that weakly equivalent cofibrant diagrams have weakly equivalent colimits.
Whenever you have a Quillen adjunction, as a pair of adjoint functors $(L,R)$, you get a pair of derived functors (L',R') which are homotopy invariant versions of $L$ and $R$ in a similar way: L'(X) is given by evaluating $L$ on a cofibrant replacement of $X$. R'(Y) is gotten by evaluating $R$ on a fibrant replacement of $Y$.
And I've rambled for long enough -- hopefully this helps a bit :).