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P 1
n
P 1 P 2
P 2
Figure 3.11.
Now we let
b j be perpendicular to −−→
n
=
a i
+
P 1 P 2
Therefore,
−−→
P 1 P 2 ·
n
=
0
and
x 2
x 1 i
+
y 2
y 1 j
·
a i
+
b j
=
0
ax 2
x 1
+
by 2
y 1
=
0
ax 2
ax 1 +
by 2
by 1 =
0
a
b =
y 1
y 2
x 2
x 1
Therefore,
a
=
y 1
y 2
and
b
=
x 2
x 1
The general form of the line equation is
y 1
y 2 x
+
x 2
x 1 y
+
c
=
0
(3.13)
c is found by substituting P 1 or P 2 in Eq. (3.13).
We now evaluate Eq. (3.13) for each boundary edge.
−−→
P 1 P 2
x 1 =
2
y 1 =
1
x 2 =
3
y 2 =
2
 
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