Graphics Reference
In-Depth Information
Figure 2.5.
Example 2.2.3.3.
L
P (x,y)
Q
P¢(x¢,y¢)
(-3,1)
A
(1,-1)
Solution.
Let
A
= (-1,0),
N
= (1/÷
-
)(2,-1), and let
P
,
Q
,
P
¢ be as shown in Figure 2.5.
Then
PA
= (-x-1,-y). Using the formulas in the definition of a reflection, it follows that
1
5
21
1
5
È
Í
˘
˙
(
)
(
)
(
)
PQ
=---
x
,
1
,
y
•
,
-
21
,
-
4
5
2
5
4
5
2
5
1
5
2
5
Ê
Ë
ˆ
¯
=-
xy
+
-
,
xy
-
+
.
Since S(
P
) =
P
+ 2
PQ
, we get that the equations for S are
3
5
4
5
8
5
x
¢=-
x
+
y
-
4
5
3
5
4
5
.
y
¢=
x
+
y
+
(2.13)
To check our answer note that S(-3,1) evaluates to (1,-1), which is what it should be.
Again see Figure 2.5. Our equations also give that S(
A
) =
A
and S(
B
) =
B
.
2.2.3.4. Proposition.
If S is the reflection about the line
L
defined by the equation
ax + by + c = 0, then
-- -
+
ax by c
ab
(
)
=
(
)
+
(
)
Sxy
,
xy
,
2
ab
,
.
(2.14)
2
2
Proof.
The proof of this formula is based on Theorem 2.2.3.1(2). We know that
N
=
(a,b) is a normal vector for
L
(although it may not be a unit vector). Therefore, if
P
= (x,y), to find the point
Q
shown in Figure 2.4, we need to find t so that
P
+ t
N
lies
on
L.
But
(
)
++
(
)
+=0
a x
+
ta
b y
tb
c
implies that
