A line in $\mathbb R^3$ (3d real space) can't be represented by a single equation. The reason for this is that a line is one-dimensional whereas space is 3-dimensional. The goal of writing a line in three-dimensional space is to eliminate two of these dimensions. To do this we need two equations: one to eliminate each extra dimension.
A little bit less abstractly, a line in $\mathbb R^3$ is just the intersection of two planes. This should be fairly intuitive, just imagine two infinite non-parallel planes and look at where they cross. So if we want to represent a line algebraically, all we need is a system of two plane equations. The equation of a plane is
$$ax+by+cz = r$$
where $a,b,c,r$ are constants. Now we can get our representation of a line by taking the solution set a system of two of these:
$$\begin{cases}a_1x+b_1y+c_1z=r_1\\a_2x+b_2y+c_2z=r_2\end{cases}$$
The solutions to this will form a line. As a concrete example, the intersection of the simple planes $x = 0$ and $y = 0$ will form the line that is the $z$ axis.
Consider two points with coordinates $(x_1,y_1),(x_2,y_2)$. Then the coordinates $(x,y)$ of any point on the line $L_{1,2}$ through $(x_1,y_1)$ and $(x_2,y_2)$ satisfy the equations
$$
\begin{cases}
x = x_1 + t (x_2 - x_1) \\
y = y_1 + t (y_2 - y_1)
\end{cases}
$$
which are called the parametric equations of $L_{1,2}$. You can get the Cartesian equation of $L_{1,2}$ by solving for $t$:
$$
y = y_1 + (x - x_1) \frac{y_2 - y_1}{x_2 - x_1}
$$
Moreover, a third point $(x_3,y_3)$ lies on $L_{1,2}$ if and only if
$$
\begin{cases}
x_3 = x_1 + t (x_2 - x_1) \\
y_3 = y_1 + t (y_2 - y_1)
\end{cases}
\leftrightarrow
\begin{cases}
x_3 - x_1 = t (x_2 - x_1) \\
y_3 - y_1 = t (y_2 - y_1)
\end{cases}
$$
that is to say, if and only if
$$
\frac{x_3 - x_1}{x_2 - x_1} = \frac{y_3 - y_1}{y_2 - y_1}
$$
i.e. if and only if the slope of $L_{1,2}$ and the slope of the line through $(x_1,y_1)$ and $(x_3,y_3)$ are the same. Note that to account for vertical lines it is better to check if
$$
(x_3 - x_1)(y_2 - y_1) = (x_2 - x_1)(y_3 - y_1)
$$
Indeed, this is apparent from the geometric picture, too. Note that here I'm going to assume that the coordinates of your points are real numbers, but that's just to fix ideas.
Fix a point $p$ in the plane. Then there is a $1:1$ correspondence between the lines through $p$ and the set $\Bbb{R} \cup \{\infty\}$, where to the vertical line corresponds the symbol $\infty$ and to every other line corresponds its slope.
Since the points $s,r,p$ are collinear (i.e. lie on the same line) if and only if the lines $L_{p,s} = \langle p,s \rangle$ and $L_{p,r} = \langle p,r \rangle$ coincide, it follows that $s,r,p$ are collinear if and only if the "numbers" corresponding to $L_{p,s}$ and $L_{p,r}$ are the same.
Best Answer
First, you only need two of those points. Let's take the first two, and make the identifications: $$(x_1,y_1) = (0,1)$$ $$ (x_2,y_2) = (2,7)$$
Then we use the slope formula: $$m = \dfrac{y_2-y_1}{x_2-x_1}$$
Finally, we plug all this into the "point slope" formula, and simplify as necessary: $$y-y_1 = m(x-x_1)$$