My textbook says, "... an equation of the line through $A$ and $B$ is $x = u + tw = u + t(v — u)$, where $t$ is a real number and $x$ denotes an arbitrary point on the line. An illustration of this is shown below:
Since $x = u + tw$ is the equation of a line, I presume that it is equivalent to the equation $y = mx + c$, where $m$ is the slope, $x$ is the x-intercept (when $y = 0$), and $c$ is the $y-$intercept (when $x = 0$). Therefore, in the equation $x = u + tw$, $w$ must represent the $x-$intercept $($when $x = 0), u$ must represent the y-intercept (when $w = 0$), and $t$ must represent the slope.
It is clear that there is a connection between the traditional equation of a line, $y = mx + c$, and the vector equation of a line, $x = u + tw = u + t(v — u)$. However, I am struggling to clearly understand the connection between the vectors and the traditional equation:
How can these vectors represent slope ($m$)?
How can these vectors represent x- and y-intercepts?
When I look at the diagram and the vector equations of a line, there does not seem to be a clear indication of how they relate to the cartesian illustration of the traditional equation of a line ($y = mx + c$). For instance, the vectors $w$ and $u$ in $x = u + tw$ would represent the $x$ and $y$ intercepts; But how can a vector represent the $x$ or $y$ intercepts? This makes no sense to me.
To provide further context, take the vector equation of a line $x = (-2,0.1)-t(6,5,2)$. In this case, $(-2,0.1) = u$ and $(6,5,2) = w$; But in what way do the vectors $u$ and $w$ represent $y$ and $x$ intercepts? This is not clear to me.
I would greatly appreciate it if the knowledgeable members of MSE could please take the time to clarify the connection between these two forms of a line.