In Jean Gallier, Basics of Affine Geometry (Ch. 2) shows an example of curved affine plan. He says: "this should dispell any idea that affine spaces are dull".
Let's consider the set ( circular paraboloid) \mathfrak{P}=\{P=(x,y,z)\in \mathbb{R}\times \mathbb{R}\times \mathbb{R}\mid x^2+y^2=z\}. Its elements are the points P=(x_1,x_2,x_1^2+x_2^2). We will organize \mathfrak{P} as a two dimensional affine space.
The vector space associated to the affine space \mathfrak{P} is \mathbb{R}^2. Tha action is the mapping
\mathfrak{P}\times \mathbb{R} \mapsto \mathfrak{P}
((x,y,x^2+y^2)\;,\;\begin{bmatrix} u\\v\end{bmatrix})\;\mapsto\;(x+u,y+v,(x+u)^2+(y+v)^2).
We also use the structural function of the affine space, \phi \;:\;\mathfrak{P}\times \mathfrak{P} \rightarrow \mathbb{R}^2,\;\;\phi(A,B)=\begin{bmatrix}b_1-a_1\\b_2-a_2 \end{bmatrix}, for A=(a_1,a_2,a_1^2+a_2^2)\;,\;B=(b_1,b_2,b_1^2+b_2^2).
In fact, \mathfrak{P} is isomorphic to \mathbb{A}_2 (materialized by the plan xOy) by projection pr_3.
Let's take the example A=(1,0,1) \in \mathfrak{P}\;,\;B=(0,1,1)\in \mathfrak{P}\;,\;\phi(A,B)=\begin{bmatrix}-1\\1 \end{bmatrix}. We choose the affine frame with the origin in point A and a linear basis \overrightarrow{e_1}, \overrightarrow{e_2}:
\mathfrak{R}=(A=(1,0,1);\overrightarrow{e_1}=\begin{bmatrix}1\\0 \end{bmatrix},\;\overrightarrow{e_2}=\begin{bmatrix}0\\1 \end{bmatrix}).
For an arbitrary point P=(x,y,x^2+y^2)\in \mathfrak{P}, the vector \phi(A,P)=\begin{bmatrix}x-1\\y\end{bmatrix}=(x-1)\overrightarrow{e_1}+y\overrightarrow{e_2}\;\;\Rightarrow\;\;P(x-1,y) are the affine Cartesian coordinates of P in the frame \mathfrak{R}. In particular in the frame \mathfrak{R} we have A=(1,0,1)=A(1-1,0)=A(0,0)\;,\;B=(0,1,1)=B(0-1,1)=B(-1,1).
Conversely, the point C(c_1,c_2) that has the (c_1,c_2) coordinates in the frame \mathfrak{R} corresponds to the point C=(c_1+1,c_2, c_1^2+c_2^2+2c_1+1) \in \mathfrak{P}.
Let us now investigate the colinear points with A and B. Points A,B,C=(c_1,c_2,c_1^2+c_2^2) are affinely dependent iff the vectors \overrightarrow{AC}\;,\;\overrightarrow{BC} are linerarly dependent. So
\overrightarrow{AC}=\phi(A,C)=\begin{bmatrix}c_1-a_1\\c_2-a_c \end{bmatrix}=\begin{bmatrix}c_1-1\\c_c \end{bmatrix}
\overrightarrow{BC}=\phi(B,C)=\begin{bmatrix}c_1-b_1\\c_2-b_2) \end{bmatrix}=\begin{bmatrix}c_1\\c_2-1 \end{bmatrix}
are linearly dependent. This means \exists\;(\lambda,\mu)\neq (0,0) so that
\lambda \cdot \overrightarrow{AC}\;=\;\mu \cdot \overrightarrow{BC} \Leftrightarrow \lambda \begin{bmatrix}c_1-1\\c_2 \end{bmatrix}=\mu \begin{bmatrix}c_1\\c_2-1 \end{bmatrix}\Leftrightarrow
\begin{cases}-\lambda+(\lambda-\mu)c_1=0\\\mu+(\lambda-\mu)c_2=0 \end{cases} \tag{1}
If \lambda=\mu, then we get from (1), -\lambda=\mu=0, case excluded. If \lambda \neq \mu, then
c_1=\frac{\lambda}{\lambda-\mu}\;,\;c_2=\frac{-\mu}{\lambda-\mu}.
Designate \alpha :=\frac{\lambda}{\lambda-\mu}\;\Rightarrow\;c_2=1-\alpha, so points "colinear" with A,\;B are
C=(\alpha,1-\alpha,1-2\alpha+2\alpha ^2) \tag{2}
which is expressed in the frame \mathfrak{R}: C(\alpha -1, 1-\alpha).
In a particular case, we get the "middle" of the bipoint \{A,B\} which is in the frame \mathfrak{R}:
\frac{1}{2}A+\frac{1}{2}B=\left (-\frac{1}{2},\frac{1}{2}\right )
to which corresponds thepoint C=\left (-\frac{1}{2}+1, \frac{1}{2},(-\frac{1}{2})^2+(\frac{1}{2})^2+2(-\frac{1}{2})+1\right )=C(\frac{1}{2},\frac{1}{2},\frac{1}{2}) \in \mathfrak{P}.
We can obtain from the points (2) the equation of the "line" AB.
In the frame \mathfrak{R} it is "x+y=0".(Attention, here x and y are the coordinates in \mathfrak{R} of a generic point M(x,y).)
In the frame Oxyz, we need to remove the parameter \alpha from (2)
\begin{cases}x=\alpha\\y=1-\alpha\\z=1-2\alpha+2\alpha^2 .\end{cases}
We obtain the equivalent equations
\begin{cases}x+y=1\\x^2+y^2=z\end{cases}
that is, the intersection of paraboloid \mathfrak{P} with the vertical plane x+y=1.
If we choose in the xOy plan the frame with the origin in C'(\frac{1}{2},\frac{1}{2})(projection of the midle of bipunct {A,B} above) and the "X" axis along the line x+y=1 we obtain(avoiding certain not too heavy calculations) the change of coordinates
\begin{cases}Y=z\\X=\frac{1}{\sqrt{2}}(1-2x) \end{cases}
in which the equation of the "line AB" is written Y=\frac{1}{2}+X^2 that is a parabola.
\blacksquare