Graphics Reference
In-Depth Information
−−→
P
4
P
1
x
1
=
1
y
1
=
2
x
2
=
2
y
2
=
1
Substituting these in Eq. (3.13), we get
x
+
y
+
c
=
0
(3.17)
Substituting the point 12 in Eq. (3.17) gives
1
+
2
+
c
=
0
∴
c
=−
3
and the line equation is
x
+
y
−
3
=
0
with a normal vector
i
+
j
Y
P
3
P
3
P
4
P
2
P
3
P
4
P
2
P
4
P
1
P
1
P
2
P
1
X
Figure 3.12.
Figure 3.12 shows the boundary with the normal vectors superimposed, and we see that all the
normal vectors point to the boundary's inside. The four expressions can now be tested with
different points, and it is the sign of the expression that confirms their space partition.
