Graphics Reference
In-Depth Information
3
2
1
2
y
=-
x
¢+
y
¢
.
We did not have to solve the first set of equations for x and y directly.
2.2.2.4. Example.
Continuing Example 2.2.2.3, suppose that we would like to find
the image
L
¢ of the line
L
defined by equation -3x + 2y = 2.
Solution.
All we have to do is substitute for x and y:
1
2
3
2
3
2
1
2
Ê
Á
ˆ
˜ +-
Ê
Á
ˆ
˜ =
-
3
x
¢ +
y
¢
2
x
¢ + ¢
y
2
.
Simplification of the terms and omitting the “¢”on the variables gives that the equa-
tion for
L
¢ is
3
2
3
2
Ê
Ë
ˆ
¯
Ê
Á
ˆ
˜
--
3
x
+-
3
+
12
y
=
.
Of course, we could also have found two points
p
and
q
on
L
and then computed the
equation for the line through the two points R(
p
) and R(
q
), but that would be more
work.
So far we have only considered rotations about the origin, but it is easy to define
rotations about an arbitrary point.
Definition.
Let
p
Œ
R
2
. The
general rotation R about
p
through an angle
q is defined
by the equation R = TR
0
T
-1
, where T is the translation that sends the origin to
p
and
R
0
is the rotation about the origin through the angle q. The point
p
is called the
center
of the rotation.
Note that a general rotation is a motion since it is a composite of motions.
2.2.2.5. Example.
To find the equations for the rotation R about the point (-3,-1)
through the angle p/3.
The translation T that sends the origin to (-3,-1) and its inverse T
-1
Solution.
are
defined by the equations
-
1
Tx x
:
¢=
-
3
T
:
x x
¢=
+
3
yy
¢=
-
1
yy
¢=
+
1
The equations for the rotation R
0
about the origin through the angle p/3 were already
computed in Example 2.2.2.3. Therefore, the equations for R = TR
0
T
-1
are
1
2
3
2
(
)
-
(
)
-
x
¢=
x
+
3
y
+
13
