Solution c01ac8eb-975c-4e9d-bfbd-67894d9cc8b6

Points: 3 pts

A line with rational slope $p/q$, where $p,q\in\mathbb Z$ and $q>0$,
that contains a lattice point $(a,b)$ also contains
\[
  (a+kq,\,b+kp),\qquad k\in\mathbb Z.
\]
Thus it contains infinitely many lattice points. This includes zero
slope, with $p=0$. Two distinct lattice points on a nonvertical line
force its slope to be rational, so an irrational-slope line can contain
at most one lattice point.

\cline{2-5}
\multicolumn{1}{c|}{} | zero | exactly one
| exactly two | more than two 

zero slope | Possible | Not Possible | Not Possible | Possible 

nonzero rational slope | Possible | Not Possible | Not Possible | Possible 

irrational slope | Possible | Possible | Not Possible | Not Possible 

The following six lines illustrate every ``Possible'' entry. Small dots
mark lattice points; larger black dots lie on the line.

Zero slope

\begin{minipage}[t]0.48\linewidth

$\displaystyle y=\tfrac12\vphantom{\dfrac{1}{2}}$

\begin{LatticeDiagram}
\draw[example line] (-2.40,0.50)--(2.40,0.50);
\end{LatticeDiagram}

Zero lattice points
\end{minipage}
\begin{minipage}[t]0.48\linewidth

$\displaystyle y=1\vphantom{\dfrac{1}{2}}$

\begin{LatticeDiagram}
\draw[example line] (-2.40,1)--(2.40,1);
\foreach \i in -2,-1,0,1,2 \LatticePoint{\i}{1}
\end{LatticeDiagram}

Infinitely many lattice points
\end{minipage}

Nonzero rational slope

\begin{minipage}[t]0.48\linewidth

$\displaystyle y=x+\tfrac12\vphantom{\dfrac{1}{2}}$

\begin{LatticeDiagram}
\draw[example line] (-2.40,-1.90)--(1.90,2.40);
\end{LatticeDiagram}

Zero lattice points
\end{minipage}
\begin{minipage}[t]0.48\linewidth

$\displaystyle y=x\vphantom{\dfrac{1}{2}}$

\begin{LatticeDiagram}
\draw[example line] (-2.40,-2.40)--(2.40,2.40);
\foreach \i in -2,-1,0,1,2 \LatticePoint{\i}{\i}
\end{LatticeDiagram}

Infinitely many lattice points
\end{minipage}

Irrational slope

\begin{minipage}[t]0.48\linewidth

$\displaystyle y=\sqrt2\,x+\tfrac14\vphantom{\dfrac{1}{2}}$

\begin{LatticeDiagram}
\draw[example line]
((-2.40-0.25)/sqrt(2),-2.40)--((2.40-0.25)/sqrt(2),2.40);
\end{LatticeDiagram}

Zero lattice points
\end{minipage}
\begin{minipage}[t]0.48\linewidth

$\displaystyle y=\sqrt2\,x\vphantom{\dfrac{1}{2}}$

\begin{LatticeDiagram}
\draw[example line]
(-2.40/sqrt(2),-2.40)--(2.40/sqrt(2),2.40);
\LatticePoint{0}{0}
\end{LatticeDiagram}

Exactly one lattice point: $(0,0)$
\end{minipage}

For $y=\sqrt2\,x+\tfrac14$, $x=0$ gives $y=\tfrac14$, while every
nonzero integer $x$ gives an irrational $y$. Hence this line contains
no lattice points. There are $2+2+2=6$ possible entries.

Final answer:
6\quad\text{(C)}

Notation definitions (retained for meaning):
\newcommand{\LatticePoint}[2]{  \filldraw[fill=black,draw=white,line width=0.4pt]
    (#1,#2) circle[radius=1.85pt];}