Showing $B(s) - \frac{s}{t}B(t)$ is normal

Question

Given that $s<t$ and $\{B(t),t\geq 0\}$ is a standard brownian motion, how would I show that $B(s) - \frac{s}{t}B(t)$ is normal? I tried doing something like $$B(s) - \frac{s}{t}B(t) = 1/t (s(B(t)+B(s)) - (t-s)B(s)) $$ which doesn't result in anything.
Also, what does showing that $\mathrm{Cov}(B(s), -\frac{s}{t}B(t)) = -s^2/t \neq 0$ mean? I thought this would mean that $B(s), -\frac{s}{t}B(t)$ are not independent, meaning their sum won't gauranteed be normal.