= Solution
One image takes $512\,\mathrm s$ if $1\,\mathrm{Hz}$ means one completed scan line per second. The slow-axis speed is $10/512=0.01953\,\mu\mathrm{m\,s^{-1}}$. Recording a wire spanning $1\,\mu\mathrm m$ along that axis therefore takes $51.2\,\mathrm s$. Meanwhile the fast-axis drift is $20/3600=0.005556\,\mu\mathrm{m\,s^{-1}}$, giving a transverse displacement of $0.2844\,\mu\mathrm m$. The apparent straight-line length is
$$
\boxed{L_{\rm app}=\sqrt{1^2+0.2844^2}=1.040\,\mu\mathrm m.}
$$
Its slow-axis projection remains $1\,\mu\mathrm m$; the overestimate refers to length along the slanted trace.
With the wire along the fast axis and drift along the slow axis, its length is acquired within one line, so \b[$L_{\rm app}\simeq1.000\,\mu\mathrm m$]. Taking a forward sweep of $1\,\mathrm s$ across $10\,\mu\mathrm m$, the wire is crossed in $0.1\,\mathrm s$ and the perpendicular displacement is only $0.556\,\mathrm{nm}$. The corresponding length change is about $1.5\times10^{-7}\,\mu\mathrm m$, far below the $19.5\,\mathrm{nm}$ pixel pitch. For the convention in which one hertz includes a forward and reverse sweep, that already negligible displacement halves. The slow-axis calculation still uses one completed line per second.
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