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Transformation (Reflection)

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

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Transformation (Reflection)

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  1. Transformation (Reflection)

  2. After see these slides, you can: Determine map or image of a curve its result of a Reflection

  3. Reflection • means to reflect • Original • shape → map • reflection axis

  4. In geometric plane, • as mirror, we use: • X axis • Y axis • Line x = m • Line y = n • Line y = x • Line y =-x

  5. Reflection to X axis ●P(x,y) ●P’(x’,y’) = P’(x,- y) x’ = x and y’ = -y Y O X

  6. Based on the picture: x’ = x y’ = -y in matrix form:

  7. So is matrixreflection to X axis

  8. 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

  9. 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)

  10. 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’

  11. 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

  12. Reflection to Y axis ●P(x,y) Y ● P’(x’,y’) = P’(-x,y) x’ = -x y’ = y X O

  13. Based on the picture: x’ = -x y’ = y in matrix form:

  14. So is matrix reflection to Y axis

  15. 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’

  16. x = -x’ and y = y’ are substituted to y = x2 – x get: y’ = (-x’)2 – (-x’) y’ = (x’)2 + x’ So, the image isy = x2 + x

  17. Reflection to line x = m ● Y ● P’(x’,y’) x’ = 2m - x y’ = y P(x,y) X O x = m

  18. 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’

  19. 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

  20. Reflection to line y = n ●P(x,y) ●P’ (x’, y’) = P’(x,2n – y) x’ = x and y’ = 2n – y Y y = n X O

  21. Example Determine 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

  22. 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

  23. 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

  24. Reflection to line y = x ●P(x,y) garis y = x Y ●P’(x’,y’) = P’(y, x) x’ = y y’ = x O X

  25. Based on the picture: x’ = y y’ = x in matrix form:

  26. So is reflection matrix to Y axis

  27. Example Image line 2x – y + 5 = 0 which reflected to line y = x is…. Solution: Reflection transformation matrix to y = x is ….

  28. Discussion Reflection transformation matrix to line y = x is

  29. x’ = y and y’ = x are substituted to 2x – y + 5 = 0 get: 2y’ – x ’ + 5 = 0 -x’ + 2y’ + 5 = 0

  30. -x’ + 2y’ + 5 = 0 multiply by (-1) → x’ – 2y’ – 5 = 0 So, the image is x – 2y + 5 = 0

  31. Reflection to line y = -x Line y = -x ●P (x,y) Y X O ● P’(x’,y’) = P’(-y,- x)

  32. Based on the picture: x’ = -y y’ = -x in matrix form:

  33. So is reflection matrix to Y axis

  34. Example 1 Image of circle x2 + y2 - 8y + 7 = 0 which reflected to line y = -x is ….

  35. Discussion: Reflection transformation matrix to line y = - x is so:

  36. → x’ = -y and y’ = -x or y = -x’ and x = -y’ Then substituted to x2 + y2 – 8y + 7 = 0

  37. 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

  38. Example 2 Coordinate image of point (-2,-3) by translation T = and followed by reflection to line y = -x is….

  39. Discussion By translation T = then point (-2,-3) → (-2 + 1, 3 – 7) → (-1,-4)

  40. Then point (-1,-4) is reflected to line y = - x

  41. → x’ = 4 and y’ = 1 So, the image is (4,1)

  42. Keep struggle and make up your mind

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