Why do the lines m and l coincide in the proof of the parallelogram rule?

I was going through this book called "A Course in Mathematics for Students of Physics Volume 1 by Paul Bamberg and Shlomo Sternberg". There in a part they said something like this:

...if we start with a point P and write
##R=P+u##
##Q=P+v##
and
##S=P+(u+v)##
then the four points
##P,Q,S,R##
lie at the four vertices of a parallelogram... The proof of this fact goes as follows. For any vector
##v=(a,b)##
and any real number t defines their product tv by
##tv=(ta,tb)##
if P is any point the set
##l=P+tv##
(as t varies over real number), is a straight line passing through P. If R is some other point, then the line
##m=R+sv##
(as s varies over real number) and l will intersect, i.e., have some point in common, if and only if there are some
##s_1##
and
##t_1##
such that,
##R+s_1v=P+t_1v##
which means that
##R=P+(t_1−s_1)v##
and hence, for every s, that
##R+sv=P+(s+t_1−s_1)v.##
This means that the lines m and l coincide. In other words, either the lines l and m coincide or they do not intersect, i.e., either they are the same or they are parallel...

Now what I don't understand is the last sentence, why m and l coincide? How can they say m and l coincide from the equation,
##R+sv=P+(s+t_1−s_1)v##
?

That seems a very complicated approach. What definition of a parallelogram are they using? In terms of vectors, that is more or less the definition of a parallelogram. Note that the vectors ##\vec u## and ##\vec v## need to be in different directions.

PeroK said:

In terms of vectors, that is more or less the definition of a parallelogram.

If we take the geometric definition of a parallelogram:

https://mathworld.wolfram.com/Parallelogram.html

And translate that into vectors, then we see (using the example in the above page) that we have:
$$\vec{AB} = \vec{DC} \ \text{and} \ \vec{AD} = \vec{BC}$$It would be much quicker simply to check that holds for the four points in your example, with ABCD replaced by PRSQ.

PeroK said:

And translate that into vectors, then we see (using the example in the above page) that we have: $$\vec{AB} = \vec{DC} \ \text{and} \ \vec{AD} = \vec{BC}$$It would be much quicker simply to check that holds for the four points in your example, with ABCD replaced by PRSQ.

And if the two lines (m and l in the OP's post) coincide, then one pair of vectors in the equations above will be zero vectors -- sort of a null parallelogram.

P+tv for all t, is the lien parallel to v and passing through P. R+sv, for all s, is the line parallel to v and passing through R. Since both lines are parallel to v, they are parallel to each other. Therefore they are either the same line or they do not meet. I.e. if they have even one point in common then they have all points in common.

The authors then assume there is one common point, namely that P+t1v = R +s1v. They then show that for every s, the point R+sv on one line, equals the point
P+(s+t1-s1)v on the other. I.e. that every point of form R+sv equals some point of form P+tv, namely the points R+sv and P+tv are equal when t = s+t1-s1. Hence the lines are equal. Actually they have shown the points of form R+sv all lie on the line consisting of points of form P+tv. To show the lines are equal they should also show the opposite, namely that all points of form P+tv also, lie on the line consisting of points of form R+sv. Maybe you can do this, unless you believe that a line which lies inside another line actually equals that line. Well you should do it anyway.