5. Given two points $A(0,1), B(6,9)$. If there is an integer point $C$ (Note: A point with both coordinates as integers is called an integer point), such that the area of $\triangle A B C$ is minimized. Then the minimum value of the area of $\triangle A B C$ is $\qquad$
5.1.

Since the slope of line $AB$ is $k=\frac{3}{4}$, the line passing through point $C$ and parallel to $AB$ can be expressed as $3 y-4 x=m$. Let the coordinates of point $C$ be $\left(x_{0}, y_{0}\right)$, where $x_{0}, y_{0}$ are integers. Therefore, we have
$$
3 y_{0}-4 x_{0}=m \text {. }
$$

Thus, $m$ is an integer.
The equation of line $AB$ is
$$
3 y-4 x-3=0 \text {. }
$$

The distance from point $C$ to line $AB$ is
$$
d=\frac{\left|3 y_{0}-4 x_{0}-3\right|}{\sqrt{3^{2}+(-4)^{2}}}=\frac{1}{5}|m-3| \text {. }
$$

When $d$ is at its minimum, the area of $\triangle ABC$ is minimized.
At this time, $m=2$ or 4.
Thus, $d_{\text {min }}=\frac{1}{5}$.
$$
S_{\triangle A B C}=\frac{1}{2} \cdot A B \cdot d_{\min }=\frac{1}{2} \times 10 \times \frac{1}{5}=1
$$