Graphics Reference
In-Depth Information
Proof.
Let (
A
¢,
B
¢,
C
¢) = M(
A
,
B
,
C
). Since M is a motion,
A C
¢¢=
AC
=
t
AB
=
t
A B
¢¢
.
The proof is divided into cases.
Case 1.
0 £ t £ 1.
Case 2.
1 < t.
Case 3.
t < 0.
In Case 1,
C
is between
A
and
B
. By Proposition 1.2.4, only
X
=¢ + ¢
A
t
AB
¢
or
X
=¢ - ¢
A
t
AB
¢
1
2
are solutions to the equation
AX
¢¢=
t
AB
¢¢
.
Of these, only
X
1
lies between
A
¢ and
B
¢. By part (1) of Theorem 2.2.1 we have that
C
¢ =
X
1
, which proves the lemma. The proofs in the other two cases are similar and
are left as exercises to the reader. Note that in Case 2
B
is between
A
and
C
and in
Case 3
A
is between
C
and
B
.
2.2.3. Lemma.
Let
L
1
and
L
2
be two distinct lines in the plane which intersect in a
point
C
. Let
P
be any point not on either of these lines. Then there exist two distinct
points
A
and
B
on
L
1
and
L
2
, respectively, so that
P
lies on the line
L
determined by
A
and
B
.
Proof.
See Figure 2.1. Let
v
1
and
v
2
be direction vectors for
L
1
and
L
2
, respectively.
These vectors are linearly independent since the lines are not parallel. Let
A
=
C
+ a
v
1
be any point on
L
1
with a > 0 and let
L
be the line determined by
P
and
A
. To find
the intersection of
L
and
L
2
, we must solve the equation
P ACv
+
s
=
+
t
2
L
2
v
2
B
L
1
v
1
A
C
P
L
Figure 2.1.
Proving Lemma 2.2.3.
