Graphics Reference
In-Depth Information
Substituting these in Eq. (3.13), we get
−
x
+
y
+
c
=
0
(3.14)
Substituting the point 21 in Eq. (3.14) gives
−
2
+
1
+
c
=
0
c
=
1
∴
and the line equation is
−
x
+
y
+
1
=
0
with a normal vector
−
i
+
j
−−→
P
2
P
3
x
1
=
3
y
1
=
2
x
2
=
2
y
2
=
3
Substituting these in Eq. (3.13), we get
−
x
−
y
+
c
=
0
(3.15)
Substituting the point 32 in Eq. (3.15) gives
−
3
−
2
+
c
=
0
∴
c
=
5
and the line equation is
−
−
+
=
x
y
5
0
with a normal vector
−
i
−
j
−−→
P
3
P
4
x
1
=
2
y
1
=
3
x
2
=
1
y
2
=
2
Substituting these in Eq. (3.13), we get
−
+
=
x
y
c
0
(3.16)
Substituting the point 23 in Eq. (3.16) gives
2
−
3
+
c
=
0
c
=
1
∴
and the line equation is
x
−
y
+
1
=
0
with a normal vector
i
−
j
