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# Geometry Section 3-6 1112

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Perpendiculars and Dist

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### Geometry Section 3-6 1112

1. 1. Section 3-6 Perpendiculars and DistanceThursday, January 5, 2012
2. 2. Essential Questions n How do you ﬁnd the distance between a point and a line? n How do you ﬁnd the distance between parallel lines?Thursday, January 5, 2012
3. 3. Vocabulary 1. Equidistant: 2. Distance Between a Point and a Line: 3. Distance Between Parallel Lines:Thursday, January 5, 2012
4. 4. Vocabulary 1. Equidistant: The distance between any two lines as measured along a perpendicular is the same; this occurs with parallel lines 2. Distance Between a Point and a Line: 3. Distance Between Parallel Lines:Thursday, January 5, 2012
5. 5. Vocabulary 1. Equidistant: The distance between any two lines as measured along a perpendicular is the same; this occurs with parallel lines 2. Distance Between a Point and a Line: The length of the segment perpendicular to the line with the point one endpoint on the segment 3. Distance Between Parallel Lines:Thursday, January 5, 2012
6. 6. Vocabulary 1. Equidistant: The distance between any two lines as measured along a perpendicular is the same; this occurs with parallel lines 2. Distance Between a Point and a Line: The length of the segment perpendicular to the line with the point one endpoint on the segment 3. Distance Between Parallel Lines: The length of the segment perpendicular to the two parallel lines with the endpoints on either of the parallel linesThursday, January 5, 2012
7. 7. Postulates & Theorems 1. Perpendicular Postulate: 2. Two Lines Equidistant from a Third:Thursday, January 5, 2012
8. 8. Postulates & Theorems 1. Perpendicular Postulate: If given a line and a point not on the line, then there exists exactly one line through the point that is perpendicular to the given line 2. Two Lines Equidistant from a Third:Thursday, January 5, 2012
9. 9. Postulates & Theorems 1. Perpendicular Postulate: If given a line and a point not on the line, then there exists exactly one line through the point that is perpendicular to the given line 2. Two Lines Equidistant from a Third: In a plane, if two lines are each equidistant from a third line, then the two lines are parallel to each otherThursday, January 5, 2012
10. 10. Example 1 The line a contains the points T(0, 0) and U(−5, 5). Find the distance between line a and the point V(1, 5).Thursday, January 5, 2012
11. 11. Example 1 The line a contains the points T(0, 0) and U(−5, 5). Find the distance between line a and the point V(1, 5). 1. Find the equation of the original lineThursday, January 5, 2012
12. 12. Example 1 The line a contains the points T(0, 0) and U(−5, 5). Find the distance between line a and the point V(1, 5). 1. Find the equation of the original line 0−5 m= 0+5Thursday, January 5, 2012
13. 13. Example 1 The line a contains the points T(0, 0) and U(−5, 5). Find the distance between line a and the point V(1, 5). 1. Find the equation of the original line 0 − 5 −5 m= = 0+5 5Thursday, January 5, 2012
14. 14. Example 1 The line a contains the points T(0, 0) and U(−5, 5). Find the distance between line a and the point V(1, 5). 1. Find the equation of the original line 0 − 5 −5 m= = = −1 0+5 5Thursday, January 5, 2012
15. 15. Example 1 The line a contains the points T(0, 0) and U(−5, 5). Find the distance between line a and the point V(1, 5). 1. Find the equation of the original line 0 − 5 −5 m= = = −1 T(0, 0) 0+5 5Thursday, January 5, 2012
16. 16. Example 1 The line a contains the points T(0, 0) and U(−5, 5). Find the distance between line a and the point V(1, 5). 1. Find the equation of the original line 0 − 5 −5 y = mx + b m= = = −1 T(0, 0) 0+5 5Thursday, January 5, 2012
17. 17. Example 1 The line a contains the points T(0, 0) and U(−5, 5). Find the distance between line a and the point V(1, 5). 1. Find the equation of the original line 0 − 5 −5 y = mx + b m= = = −1 T(0, 0) 0+5 5 y = −xThursday, January 5, 2012
18. 18. Example 1 The line a contains the points T(0, 0) and U(−5, 5). Find the distance between line a and the point V(1, 5). 1. Find the equation of the original line 0 − 5 −5 y = mx + b m= = = −1 T(0, 0) 0+5 5 y = −x2. Find the equation of the perpendicular line through the other pointThursday, January 5, 2012
19. 19. Example 1 The line a contains the points T(0, 0) and U(−5, 5). Find the distance between line a and the point V(1, 5). 1. Find the equation of the original line 0 − 5 −5 y = mx + b m= = = −1 T(0, 0) 0+5 5 y = −x2. Find the equation of the perpendicular line through the other point m =1Thursday, January 5, 2012
20. 20. Example 1 The line a contains the points T(0, 0) and U(−5, 5). Find the distance between line a and the point V(1, 5). 1. Find the equation of the original line 0 − 5 −5 y = mx + b m= = = −1 T(0, 0) 0+5 5 y = −x2. Find the equation of the perpendicular line through the other point m =1 V(1, 5)Thursday, January 5, 2012
21. 21. Example 1 The line a contains the points T(0, 0) and U(−5, 5). Find the distance between line a and the point V(1, 5). 1. Find the equation of the original line 0 − 5 −5 y = mx + b m= = = −1 T(0, 0) 0+5 5 y = −x2. Find the equation of the perpendicular line through the other point m =1 y − y1 = m(x − x1 ) V(1, 5)Thursday, January 5, 2012
22. 22. Example 1 The line a contains the points T(0, 0) and U(−5, 5). Find the distance between line a and the point V(1, 5). 1. Find the equation of the original line 0 − 5 −5 y = mx + b m= = = −1 T(0, 0) 0+5 5 y = −x2. Find the equation of the perpendicular line through the other point m =1 y − y1 = m(x − x1 ) V(1, 5) y − 5 = 1(x − 1)Thursday, January 5, 2012
23. 23. Example 1 The line a contains the points T(0, 0) and U(−5, 5). Find the distance between line a and the point V(1, 5). 1. Find the equation of the original line 0 − 5 −5 y = mx + b m= = = −1 T(0, 0) 0+5 5 y = −x2. Find the equation of the perpendicular line through the other point m =1 y − y1 = m(x − x1 ) y − 5 = x −1 V(1, 5) y − 5 = 1(x − 1)Thursday, January 5, 2012
24. 24. Example 1 The line a contains the points T(0, 0) and U(−5, 5). Find the distance between line a and the point V(1, 5). 1. Find the equation of the original line 0 − 5 −5 y = mx + b m= = = −1 T(0, 0) 0+5 5 y = −x2. Find the equation of the perpendicular line through the other point m =1 y − y1 = m(x − x1 ) y − 5 = x −1 V(1, 5) y − 5 = 1(x − 1) y = x+ 4Thursday, January 5, 2012
25. 25. Example 1 3. Solve the system of these two equations.Thursday, January 5, 2012
26. 26. Example 1 3. Solve the system of these two equations. ⎧y = − x ⎨ ⎩y = x + 4Thursday, January 5, 2012
27. 27. Example 1 3. Solve the system of these two equations. ⎧y = − x −x = x + 4 ⎨ ⎩y = x + 4Thursday, January 5, 2012
28. 28. Example 1 3. Solve the system of these two equations. ⎧y = − x −x = x + 4 ⎨ −2x = 4 ⎩y = x + 4Thursday, January 5, 2012
29. 29. Example 1 3. Solve the system of these two equations. ⎧y = − x −x = x + 4 ⎨ −2x = 4 ⎩y = x + 4 x = −2Thursday, January 5, 2012
30. 30. Example 1 3. Solve the system of these two equations. ⎧y = − x −x = x + 4 y = −(−2) ⎨ −2x = 4 ⎩y = x + 4 x = −2Thursday, January 5, 2012
31. 31. Example 1 3. Solve the system of these two equations. ⎧y = − x −x = x + 4 y = −(−2) = 2 ⎨ −2x = 4 ⎩y = x + 4 x = −2Thursday, January 5, 2012
32. 32. Example 1 3. Solve the system of these two equations. ⎧y = − x −x = x + 4 y = −(−2) = 2 ⎨ −2x = 4 ⎩y = x + 4 2 = −2 + 4 x = −2Thursday, January 5, 2012
33. 33. Example 1 3. Solve the system of these two equations. ⎧y = − x −x = x + 4 y = −(−2) = 2 ⎨ −2x = 4 ⎩y = x + 4 2 = −2 + 4 x = −2 (−2,2)Thursday, January 5, 2012
34. 34. Example 1 4. Use the distance formula utilizing this point on the line and the point not on the line.Thursday, January 5, 2012
35. 35. Example 1 4. Use the distance formula utilizing this point on the line and the point not on the line. (1, 5), (−2, 2)Thursday, January 5, 2012
36. 36. Example 1 4. Use the distance formula utilizing this point on the line and the point not on the line. (1, 5), (−2, 2) d = (x2 − x1 ) + (y2 − y1 ) 2 2Thursday, January 5, 2012
37. 37. Example 1 4. Use the distance formula utilizing this point on the line and the point not on the line. (1, 5), (−2, 2) d = (x2 − x1 ) + (y2 − y1 ) = (−2 − 1) + (2 − 5) 2 2 2 2Thursday, January 5, 2012
38. 38. Example 1 4. Use the distance formula utilizing this point on the line and the point not on the line. (1, 5), (−2, 2) d = (x2 − x1 ) + (y2 − y1 ) = (−2 − 1) + (2 − 5) 2 2 2 2 = (−3) + (−3) 2 2Thursday, January 5, 2012
39. 39. Example 1 4. Use the distance formula utilizing this point on the line and the point not on the line. (1, 5), (−2, 2) d = (x2 − x1 ) + (y2 − y1 ) = (−2 − 1) + (2 − 5) 2 2 2 2 = (−3) + (−3) = 9 + 9 2 2Thursday, January 5, 2012
40. 40. Example 1 4. Use the distance formula utilizing this point on the line and the point not on the line. (1, 5), (−2, 2) d = (x2 − x1 ) + (y2 − y1 ) = (−2 − 1) + (2 − 5) 2 2 2 2 = (−3) + (−3) = 9 + 9 = 18 2 2Thursday, January 5, 2012
41. 41. Example 1 4. Use the distance formula utilizing this point on the line and the point not on the line. (1, 5), (−2, 2) d = (x2 − x1 ) + (y2 − y1 ) = (−2 − 1) + (2 − 5) 2 2 2 2 = (−3) + (−3) = 9 + 9 = 18 ≈ 4.24 2 2Thursday, January 5, 2012
42. 42. Example 1 4. Use the distance formula utilizing this point on the line and the point not on the line. (1, 5), (−2, 2) d = (x2 − x1 ) + (y2 − y1 ) = (−2 − 1) + (2 − 5) 2 2 2 2 = (−3) + (−3) = 9 + 9 = 18 ≈ 4.24 units 2 2Thursday, January 5, 2012
43. 43. Example 2 Find the distance between the parallel lines m and n with the following equations. y = 2x + 3 y = 2x − 1Thursday, January 5, 2012
44. 44. Example 2 Find the distance between the parallel lines m and n with the following equations. y = 2x + 3 y = 2x − 1 1. Find the equation of the perpendicular line.Thursday, January 5, 2012
45. 45. Example 2 Find the distance between the parallel lines m and n with the following equations. y = 2x + 3 y = 2x − 1 1. Find the equation of the perpendicular line. y = mx + bThursday, January 5, 2012
46. 46. Example 2 Find the distance between the parallel lines m and n with the following equations. y = 2x + 3 y = 2x − 1 1. Find the equation of the perpendicular line. y = mx + b 1 m = − ,(0,3) 2Thursday, January 5, 2012
47. 47. Example 2 Find the distance between the parallel lines m and n with the following equations. y = 2x + 3 y = 2x − 1 1. Find the equation of the perpendicular line. y = mx + b 1 m = − ,(0,3) 2 1 y = − x+3 2Thursday, January 5, 2012
48. 48. Example 2 2. Find the intersection of the perpendicular line and the other parallel line using a system.Thursday, January 5, 2012
49. 49. Example 2 2. Find the intersection of the perpendicular line and the other parallel line using a system. ⎧ y = 2x − 1 ⎪ ⎨ 1 ⎪y = − x + 3 ⎩ 2Thursday, January 5, 2012
50. 50. Example 2 2. Find the intersection of the perpendicular line and the other parallel line using a system. ⎧ y = 2x − 1 ⎪ ⎨ 1 ⎪y = − x + 3 ⎩ 2 1 2x − 1 = − x + 3 2Thursday, January 5, 2012
51. 51. Example 2 2. Find the intersection of the perpendicular line and the other parallel line using a system. ⎧ y = 2x − 1 ⎪ ⎨ 1 ⎪y = − x + 3 ⎩ 2 1 2x − 1 = − x + 3 2 5 x= 4 2Thursday, January 5, 2012
52. 52. Example 2 2. Find the intersection of the perpendicular line and the other parallel line using a system. ⎧ y = 2x − 1 ⎪ ⎨ 1 ⎪y = − x + 3 ⎩ 2 1 2x − 1 = − x + 3 2 5 x= 4 x = 1.6 2Thursday, January 5, 2012
53. 53. Example 2 2. Find the intersection of the perpendicular line and the other parallel line using a system. ⎧ y = 2x − 1 ⎪ y = 2(1.6) − 1 ⎨ 1 ⎪y = − x + 3 ⎩ 2 1 2x − 1 = − x + 3 2 5 x= 4 x = 1.6 2Thursday, January 5, 2012
54. 54. Example 2 2. Find the intersection of the perpendicular line and the other parallel line using a system. ⎧ y = 2x − 1 ⎪ y = 2(1.6) − 1 ⎨ 1 ⎪y = − x + 3 y = 2.2 ⎩ 2 1 2x − 1 = − x + 3 2 5 x= 4 x = 1.6 2Thursday, January 5, 2012
55. 55. Example 2 2. Find the intersection of the perpendicular line and the other parallel line using a system. ⎧ y = 2x − 1 1 ⎪ y = 2(1.6) − 1 y = − (1.6) − 1 ⎨ 1 2 ⎪y = − x + 3 y = 2.2 ⎩ 2 1 2x − 1 = − x + 3 2 5 x= 4 x = 1.6 2Thursday, January 5, 2012
56. 56. Example 2 2. Find the intersection of the perpendicular line and the other parallel line using a system. ⎧ y = 2x − 1 1 ⎪ y = 2(1.6) − 1 y = − (1.6) − 1 ⎨ 1 2 ⎪y = − x + 3 y = 2.2 ⎩ 2 y = 2.2 1 2x − 1 = − x + 3 2 5 x= 4 x = 1.6 2Thursday, January 5, 2012
57. 57. Example 2 2. Find the intersection of the perpendicular line and the other parallel line using a system. ⎧ y = 2x − 1 1 ⎪ y = 2(1.6) − 1 y = − (1.6) − 1 ⎨ 1 2 ⎪y = − x + 3 y = 2.2 ⎩ 2 y = 2.2 1 2x − 1 = − x + 3 (1.6, 2.2) 2 5 x= 4 x = 1.6 2Thursday, January 5, 2012
58. 58. Example 23. Use the new point and original y-intercept you chose in step 2 in the distance formula.Thursday, January 5, 2012
59. 59. Example 23. Use the new point and original y-intercept you chose in step 2 in the distance formula. (0, 3), (1.6, 2.2)Thursday, January 5, 2012
60. 60. Example 23. Use the new point and original y-intercept you chose in step 2 in the distance formula. (0, 3), (1.6, 2.2) d = (x2 − x1 ) + (y2 − y1 ) 2 2Thursday, January 5, 2012
61. 61. Example 23. Use the new point and original y-intercept you chose in step 2 in the distance formula. (0, 3), (1.6, 2.2) d = (x2 − x1 ) + (y2 − y1 ) = (1.6 − 0) + (2.2 − 3) 2 2 2 2Thursday, January 5, 2012
62. 62. Example 23. Use the new point and original y-intercept you chose in step 2 in the distance formula. (0, 3), (1.6, 2.2) d = (x2 − x1 ) + (y2 − y1 ) = (1.6 − 0) + (2.2 − 3) 2 2 2 2 = (1.6) + (−0.8) 2 2Thursday, January 5, 2012
63. 63. Example 23. Use the new point and original y-intercept you chose in step 2 in the distance formula. (0, 3), (1.6, 2.2) d = (x2 − x1 ) + (y2 − y1 ) = (1.6 − 0) + (2.2 − 3) 2 2 2 2 = (1.6) + (−0.8) = 2.56 + .64 2 2Thursday, January 5, 2012
64. 64. Example 23. Use the new point and original y-intercept you chose in step 2 in the distance formula. (0, 3), (1.6, 2.2) d = (x2 − x1 ) + (y2 − y1 ) = (1.6 − 0) + (2.2 − 3) 2 2 2 2 = (1.6) + (−0.8) = 2.56 + .64 2 2 = 3.2Thursday, January 5, 2012
65. 65. Example 23. Use the new point and original y-intercept you chose in step 2 in the distance formula. (0, 3), (1.6, 2.2) d = (x2 − x1 ) + (y2 − y1 ) = (1.6 − 0) + (2.2 − 3) 2 2 2 2 = (1.6) + (−0.8) = 2.56 + .64 2 2 = 3.2 ≈ 1.79Thursday, January 5, 2012
66. 66. Example 23. Use the new point and original y-intercept you chose in step 2 in the distance formula. (0, 3), (1.6, 2.2) d = (x2 − x1 ) + (y2 − y1 ) = (1.6 − 0) + (2.2 − 3) 2 2 2 2 = (1.6) + (−0.8) = 2.56 + .64 2 2 = 3.2 ≈ 1.79 unitsThursday, January 5, 2012
67. 67. Example 3 You try it out! Refer to the process in example 1. Line h contains the points E(2, 4) and F(5, 1). Find the distance between line h and the point G(1, 1).Thursday, January 5, 2012
68. 68. Example 3 You try it out! Refer to the process in example 1. Line h contains the points E(2, 4) and F(5, 1). Find the distance between line h and the point G(1, 1). Solution:Thursday, January 5, 2012
69. 69. Example 3 You try it out! Refer to the process in example 1. Line h contains the points E(2, 4) and F(5, 1). Find the distance between line h and the point G(1, 1). Solution: d= 8Thursday, January 5, 2012
70. 70. Example 3 You try it out! Refer to the process in example 1. Line h contains the points E(2, 4) and F(5, 1). Find the distance between line h and the point G(1, 1). Solution: d = 8 ≈ 2.83Thursday, January 5, 2012
71. 71. Example 3 You try it out! Refer to the process in example 1. Line h contains the points E(2, 4) and F(5, 1). Find the distance between line h and the point G(1, 1). Solution: d = 8 ≈ 2.83 unitsThursday, January 5, 2012
72. 72. Check Your Understanding Review problems #1-8 on p. 218Thursday, January 5, 2012
73. 73. Problem SetThursday, January 5, 2012
74. 74. Problem Set p. 218 #13-33 odd, 53, 59, 63 “I’m a great believer in luck, and I ﬁnd the harder I work the more I have of it.” - Thomas JeffersonThursday, January 5, 2012