transformation (reflection)
DESCRIPTION
Transformation (Reflection). After see these slides, you can: Determine map or image of a curve its result of a Reflection. Reflection means to reflect Original shape → map reflection axis. In geometric plane, as mirror, we use: X axis - PowerPoint PPT PresentationTRANSCRIPT
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TransformatiTransformationon
(Reflection)(Reflection)
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After see these slides, you can:
Determine map or image of a curve its result of a
Reflection
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Reflection means to reflect
Original
shape → map
reflection axis
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In geometric plane,
as mirror, we use:X axisY axis
Line x = mLine y = nLine y = xLine y =-x
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Reflection to X axis ●P(x,y)
●P’(x’,y’) = P’(x,- y)
x’ = x and y’ = -y
XO
Y
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Based on the picture:
x’ = x
y’ = -y
in matrix form:
y
x
y
x
10
01
'
'
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So
is matrix reflection to X axis
10
01
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Example 1
Known triangle ABC with
coordinate A(2,0), B(0,-5) and
C(-3,1). Find the image coordinate of triangle ABC if it is reflected to X axis
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Discussion
Reflection to X axis
P (x, y) → P’(-x, y)
The images are:
A(2,0) is A’(-2,0)
B(0,-5) is B’(0,-5)
C(-3,1) is C’(3,1)
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Example 2
The image line 3x – 2y + 5 = 0 by
reflection to X axis is….
Solution:
By reflection to X axis
so: x’ = x → x = x’
y’ = -y → y = -y’
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x = x’ and y = -y’
Is substituted to curve
3x – 2y + 5 = 0
get: 3x’ – 2(-y’) + 5 = 0
3x’ + 2y’ + 5 = 0
So, the image is 3x + 2y + 5 = 0
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Reflection to Y axis
●P(x,y)
●
O
Y
P’(x’,y’)= P’(-x,y)
x’ = -x y’ = y
X
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Based on the picture:
x’ = -x
y’ = y
in matrix form:
y
x
y
x
10
01
'
'
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So
is matrix reflection to Y axis
10
01
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Example
Determine map of curve y = x2 – x
by reflection to Y axis.
Solution:
By reflection to Y axis
then: x’ = -x → x = -x’
y’ = y → y = y’
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x = -x’ and y = y’
are substituted to y = x2 – x
get: y’ = (-x’)2 – (-x’)
y’ = (x’)2 + x’
So, the image is y = x2 + x
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Reflection to line x = m
● ●
O
YP’(x’,y’)x’ = 2m - xy’ = y
X
x = m
P(x,y)
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Example
Determine image curve y2 = x – 5
by reflection to line x = 3.
Solution:
by reflection to line x = 3
then: x’ = 2m - x → x = 2.3 - x’ = 6 –x’
y’ = y → y = y’
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x = 6 – x’ and y = y’ are
substituted to y2 = x - 5
get: (y’)2 = (6 – x’) – 5
(y’)2 = 1 – x’
So, the image is y2 = 1 - x
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Reflection to line y = n ●P(x,y)
●P’ (x’, y’) = P’(x,2n – y)
x’ = x and y’ = 2n – y
XO
Y
y = n
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ExampleDetermine the image curve
x2 + y2 = 4
by reflection to line y = -3.
Solution: by reflection to line y = - 3,
so: x’ = x
y’ = 2n - y
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reflection to line y = - 3
then: x’ = x x = x’
y’ = 2n – y
y’ = 2(-3) – y
y’ = - 6 – y y = -y’ – 6
is substituted to x2 + y2 = 4
(x’)2 + (-y’ – 6)2 = 4
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then substituted to x2 + y2 = 4
(x’)2 + (-y’ – 6)2 = 4
(x’)2 +((-y’)2 + 12y’ + 36) – 4 = 0
So, the image is:
x2 + y2 + 12y + 32 = 0
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Reflection to line y = x ●P(x,y)
garis y = x
XO
Y
●P’(x’,y’) = P’(y, x)
x’ = y
y’ = x
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Based on the picture:
x’ = y
y’ = x
in matrix form:
y
x
y
x
01
10
'
'
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So
is reflection matrix to Y axis
01
10
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ExampleImage line 2x – y + 5 = 0
which reflected to line y = x is….
Solution:
Reflection transformation matrix to
y = x is ….
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DiscussionReflection transformation matrix to
line y = x is
01
10
x
y
y
x
y
x
01
10
'
'
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x
y
y
x
y
x
01
10
'
'
x’ = y and y’ = x
are substituted to 2x – y + 5 = 0
get: 2y’ – x ’ + 5 = 0
-x’ + 2y’ + 5 = 0
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-x’ + 2y’ + 5 = 0
multiply by (-1) → x’ – 2y’ – 5 = 0
So, the image is x – 2y + 5 = 0
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Reflection to line y = -x
XO
Y
●P’(x’,y’) = P’(-y,- x)
Line y = -x ●P (x,y)
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Based on the picture:
x’ = -y
y’ = -x
in matrix form:
y
x
y
x
01
10
'
'
3333
So
is reflection matrix to Y axis
01
10
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Example 1
Image of circle
x2 + y2 - 8y + 7 = 0
which reflected to line y = -x is ….
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Discussion:Reflection transformation matrix
to line y = - x is
so:
01
10
y
x
y
x
01
10
'
'
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x
y
y
x
y
x
01
10
'
'
→ x’ = -y and y’ = -x
or y = -x’ and x = -y’
Then substituted to
x2 + y2 – 8y + 7 = 0
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x = -y’ and y = -x’ are substituted
to x2 + y2 – 8y + 7 = 0
→ (-y’)2 + (-x)2 – 8(-x) + 7 = 0
(y’)2 + (x’)2 + 8x + 7 = 0
(x’)2 + (y’)2 + 8x + 7 = 0
So, the image is
x2 + y2 + 8x + 7 = 0
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Example 2Coordinate image of point (-2,-3)
by translation T =
and followed by reflection to
line y = -x is….
7
1
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Discussion
By translation T =
then point (-2,-3) → (-2 + 1, 3 – 7)
→ (-1,-4)
7
1
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Then point (-1,-4) is
reflected to line y = - x
y
x
y
x
01
10
'
'
4
1
01
10
'
'
y
x
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→ x’ = 4 and y’ = 1
So, the image is (4,1)
4
1
01
10
'
'
y
x
1
4
)4.(0)1)(1(
)4)(1()1.(0
'
'
y
x
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