Subsection 1.2 Lines
Theorem 1 Every pair of distinct points p 6= q in R2 determines a line.
First, choose x0 = p and v = q − p. Then
p
=
x0 + 0 · v
q
=
x0 + 1 · v
so the line x = p + t(q − p) contains the two points p, q. Second, we have to
prove that this is the only line passing through p and q. Let x = x00 + tv 0 be
any such line passing through p and q. Then there exist number r, s such that
p
=
x00 + rv 0
q
=
x00 + sv 0
For any particular point x = x00 + tv 0 , we have to show that it is also part of the
line x = p + t(p − q). Using
x00 − p
= −rv 0
(s − r)v 0
= q−p
we express x in terms of p and q:
x =
x00 + tv 0 = p + x00 − p + tv 0 = p − rv 0 +
= p+
t
(q − p)
s−r
r(s − r) + t
(q − p)
s−r
Note that when we divided by r − s, we could do this since s − r = 0 would
mean p = q, which is not the case by the hypothesis that p and q are distinct.
The last equation shows that x is indeed on the line x = p + t(q − p).