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Transformation ( Translation, Reflection, Rotation , and Dilatation)

Transformation ( Translation, Reflection, Rotation , and Dilatation). After see these slides you can: Determine map or image of result curve of a Translation , Reflection, Rotation , or Dilatation. Transformation To move a point or shape on a plane can be done by transformation.

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Transformation ( Translation, Reflection, Rotation , and Dilatation)

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  1. Transformation (Translation, Reflection, Rotation, and Dilatation)

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

  3. Transformation To move a point or shape on a plane can be done by transformation. Transformation T on a plane is ‘mapping’ every P point on plane become P’ on the same plane. P’ is called image or map of P

  4. Kind of Transformation: a. Translation b. Reflection c. Rotation d. Dilatation *) will be discussed today

  5. Translation • means friction

  6. If translation T = mapping P (x, y) to P´ (x’, y’) then x’ = x + a and y’ = y + b Is written into matrix:

  7. Example 1 Known triangle OAB with coordinate O(0,0), A(3,0), and B(3,5). Find the image coordinate of triangle OAB if it is translated by T =

  8. Discussion: (0,0) → (0 + 1, 0 + 3) 0’(1,3) (3,0) → (3 + 1, 0 + 3) A’(4,3) (3,5) → (3 + 1, 5 + 3) B’(4,8) y X O

  9. Example 2 Image of circle equation x2 + y2 = 25 by translation T = is….

  10. ● Discussion: P (-1,3) ● X

  11. Because translation T = so • x’ = x – 1 → x = x’ + 1.….(1) • y’ = y + 3 → y = y’ – 3…..(2) • and (2) are substituted into • x2 + y2 = 25 can be found • (x’ + 1)2 + (y’ – 3)2 = 25; • Then the image is: (x + 1)2 + (y – 3)2 = 25

  12. Example 3 By a translation, map of point (1,-5) is (7,-8). The image curve of y = x2 + 4x – 12 by that translation is….

  13. Discussion If translation T = Image point (1,-5) by translation T is (1 + a, -5 + b) = (7,-8) 1+ a = 7 → a = 6 -5+ b = -8 → b = -3

  14. a = 6 and b = -3 so the translation is T = Because T = Then x’ = x + 6 → x = x’ – 6 y’ = y – 3 → y = y’ + 6

  15. x = x’ – 6 and y = y’ + 3 are substituted into y = x2 + 4x – 12 y’ + 3 = (x’ – 6)2 + 4(x’ – 6) – 12 y’ + 3 = (x’)2 – 12x’ + 36 + 4x’ - 24 -12 y’ = (x’)2 – 8x’ – 3 So, the image is: y = x2 – 8x – 3

  16. Rotation • means move by circular path • determine by • centrum and angular magnitude

  17. Rotation with centre O(0,0) • P(x,y) is rotated as  counter • clockwise with centre O(0,0) and • can be found image P’(x’,y’) • so: x’ = xcos - ysin • y’ = xsin + ycos

  18. If angular angle  = ½π (the rotation is signed with R½π) then x’ = - y andy’ = x In matrix: so R½π=

  19. Example 1 • Equation of image line • x + y = 6 after it is rotated • in centre coordinate with angular angle +90o, is….

  20. Discussion: • R+90omeans: x’ = -y → y = -x’ • y’ = x → x = y’ • are substituted into: x + y = 6 • y’ + (-x’) = 6 • y’ – x’ = 6 → x’ – y’ = - 6 • So the image line x + y = 6 is: x – y = - 6

  21. Example 2 • Equation of image line • 2x - y + 6 = 0 after it is rotated • in centre coordinate with • angular angle -90o, is….

  22. Discussion: • R-90omeans: • x’ = xcos(-90) – ysin(-90) • y’ = xsin(-90) + ycos(-90) • x’ = 0 – y(-1) = y • y’ = x(-1) + 0 = -x’ or • with matrix:

  23. R-90omeans: x’ = y → y = x’ y’ = -x → x = -y’ substituted into:2x - y + 6 = 0 2(-y’) - x’ + 6 = 0 -2y’ – x’ + 6 = 0 x’ + 2y’ – 6 = 0 So, the image line2x - y + 6 = 0is: x + y – 6 = 0

  24. If angular angle  = π (the rotation is signed H) with x’ = - x dany’ = -y in matrix: so H=

  25. Example • Equation of image parabola • y = 3x2 – 6x + 1 after it is rotated in centre coordinate with angular angle +180o, is….

  26. Discussion • Hmeans: x’ = -x → x = -x’ • y’ = -y → y = -y’ • substituted into: y = 3x2 – 6x + 1 • -y’= 3(-x’)2 – 6(-x’) + 1 • -y’= 3(x’)2 + 6x + 1 • (multiplied by -1) • So, the image parabola y = 3x2 – 6x + 1 : • is y = - 3x2 – 6x - 1

  27. Dilatation • isa transformation that change the size (bigger or smaller) ashapewithout changing its form.

  28. Dilatation centre O(0,0) and k scale factor If P (x, y) is dilatated into O(0,0) and k scale factor the image is P’ (x’, y’) so x’ = kx and y’ = ky and it is signed with D[O, k]

  29. Example • Line 2x – 3y = 6 intersect X axis at A and intersect Y axis at B. • Because of the dilatation D[O,-2], • A become A’ and B become B’. • Find the area of OA’B’

  30. Discussion • Line 2x – 3y = 6 intersect X axis at • A(3,0), and through Y axis at B(0,2) • because the dilatation D[O,-2] then • A’ (kx, ky)→ A’(-6,0) and • B’ (kx, ky) → B’(0,-4)

  31. Y B • 4 X • A O • 6 A’(-6,0), B’(0,-4), and O(0,0) create a triangle as in the picture: then the area = ½ x OA’ x OB’ = ½ x 6 x 4 = 12

  32. Dilatation with centre P(a,b) and k scale factor The image is: x’ = k (x – a) + a and y’ = k (y – b) + b signed with [P (a, b) ,k]

  33. Example • A(-5,13) is dilatated • with [P,⅔] become A’. • If coordinate P(1,-2), then • coordinate A’ is….

  34. Discussion • A(x,y) A’(x’,y’) • x’ = k(x – a) + a • y’ = k(y – b) + b • A(-5,13) A’(x’ y’) • x’ = k(x – a) + a • y’ = k(y – b) + b [P(a,b) ,k] [P(1,-2),⅔]

  35. A(-5,13) A’(x’ y’) x’ = ⅔(-5 – 1) + 1 = -3 y’ = ⅔(13 – (-2)) + (-2) = 8 So, the coordinate A’ is (-3,8) [P(1,-2),⅔]

  36. Inverse Transformation • To determine image of curve • by transformation • written in matrix form, we can use • inverse transformation

  37. Example • Map of line x – 2y + 5 = 0 • by transformation that is stated • with matrix is….

  38. Discussion • A(x,y) A’(x’ y’) • Remember: A = BX so X = B-1.A

  39. get: x = 3x’ – y’ and y = -2x’ + y’

  40. x = 3x’ – y’ and y = -2x’ + y’ are substituted into x – 2y + 5 = 0 3x’ – y’ – 2(-2x’ + y’) + 5 = 0 3x’ – y’ + 4x’ – 2y’ + 5 = 0 7x’ – 3y’ + 5 = 0 So, the image is: 7x – 3y + 5 = 0

  41. Keep analytic and faithful

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