Graphics Reference
In-Depth Information
Y
P
2
P
3
P
i
P
1
P
4
X
Fig. 10.21.
Two line segments with their associated position vectors.
Therefore, the parameters
s
and
t
are given by
s
=
(
P
1
−
P
3
)+
t
(
P
2
−
P
1
)
(
P
4
−
P
3
)
t
=
(
P
3
−
P
1
)+
s
(
P
4
−
P
3
)
(10.46)
(
P
2
−
P
1
)
From (10.46) we can write
t
=
(
x
3
−
x
1
)+
s
(
x
4
−
x
3
)
(
x
2
−
x
1
)
t
=
(
y
3
−
y
1
)+
s
(
y
4
−
y
3
)
(10.47)
(
y
2
−
y
1
)
which yields
t
=
x
1
(
y
4
−
y
3
)+
x
3
(
y
1
−
y
4
)+
x
4
(
y
3
−
y
1
)
(
y
2
−
y
1
)(
x
4
−
x
3
)
−
(
x
2
−
x
1
)(
y
4
−
y
3
)
and similarly,
s
=
x
1
(
y
3
−
y
2
)+
x
2
(
y
3
−
y
1
)+
x
3
(
y
2
−
y
1
)
(10.48)
(
y
4
−
y
3
)(
x
2
−
x
1
)
−
(
x
4
−
x
3
)(
y
2
−
y
1
)
Let's test (10.48) with two examples to illustrate how this equation can be
used in practice. The first example will demonstrate an intersection condition,
and the second demonstrates a touching condition.
•
Example 1
. Figure 10.22a shows two line segments intersecting, with an
obvious intersection point of (1.5, 0.0). The coordinates of the line segments
are
(
x
1
,y
1
)=(1
,
0)
(
x
2
,y
2
)=(2
,
0)
(
x
3
,y
3
)=(1
.
5
,
−
1
.
0)
(
x
4
,y
4
)=(1
.
5
,
1
.
0)