Transformations - Transformations and Projections in Computer Graphics - page 14

must therefore satisfy

a x + x ∗
2

+ b y + y ∗
2

+ c =0 . (1.11)

Equations (1.10) and (1.11) can easily be solved for x ∗ and y ∗ .Thesolutionsare
P ∗ =( x ∗ ,y ∗ )= x

2 a ( ax + by + c )
a 2 + b 2

2 b ( ax + by + c )
a 2 + b 2

− ,y −

= ( b 2

.

a 2 ) x 2 abx +( a 2 b 2 ) y − − 2 aby − 2 ac

, −

− − 2 bc

(1.12)

a 2 + b 2 a 2 + b 2

Equation (1.12) is easy to verify intuitively for vertical and for horizontal lines.
When b is zero, the line becomes the vertical line x =

− c/a and Equation (1.12) reduces

to

P ∗ =( x ∗ ,y ∗ )= x ,y =

a ,y .

2 a ( ax + c )
a 2

2 c

− − x −

When a =0,thelineisthehorizontal y = − c/b ,andthesameequationreducesto

P ∗ =( x ∗ ,y ∗ )= x, y

= x, .

2 b ( by + c )
b 2

2 c
b

− − y −

The transformation matrix for reflection about an arbitrary line ax + by + c =0is
directly obtained from Equation (1.12)

⎛ ⎞

b 2 a 2

− − 2 ab 0

⎝ ⎠ .

a 2 b 2 T = − 2 ab − 0 (1.13)

1
a 2 + b 2

− 2 ac − 2 bc

Its determinant is

det T = ( b 2

a 2 )( a 2 b 2 ) 4 a 2 b 2 a 4 +2 a 2 b 2 + b 4

a 2 + b 2

− − −

( a 2 + b 2 ) ,

= − = −

a 2 + b 2

which equals 1 (pure reflection) for lines expressed in the standard form (defined as

thecasewhere a 2 + b 2 =1).

−

Exercise 1.16: Use Equation (1.12) to obtain the transformation rule for reflection
about a line that passes through the origin.

We turn now to the product of two reflections about the two arbitrary lines L 1 :
ax + by + c =0and L 2 : dx + ey + f = 0 (Figure 1.7a). This product can be calculated
from Equation (1.13) as the matrix product
⎛ ⎞ ⎛ ⎞

b 2 a 2 e 2 d 2

− − 2 ab 0 − − 2 de 0

⎝ ⎠ ⎝ ⎠ ,

− 2 ab a 2 − b 2 0 − 2 de d 2 − e 2 0

1
a 2 + b 2

1
d 2 + e 2

− 2 ac − 2 bc − 2 df − 2 ef