1. Section 3-6
Perpendiculars and Distance
Thursday, January 5, 2012
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?
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3. Vocabulary
1. Equidistant:
2. Distance Between a Point and a Line:
3. Distance Between Parallel Lines:
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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. 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. 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 lines
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7. Postulates & Theorems
1. Perpendicular Postulate:
2. Two Lines Equidistant from a Third:
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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:
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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 other
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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).
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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 line
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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+5
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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 5
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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 5
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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 5
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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 5
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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 = −x
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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 = −x
2. Find the equation of the perpendicular line through the
other point
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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 = −x
2. Find the equation of the perpendicular line through the
other point
m =1
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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 = −x
2. Find the equation of the perpendicular line through the
other point
m =1
V(1, 5)
Thursday, January 5, 2012
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 = −x
2. Find the equation of the perpendicular line through the
other point
m =1 y − y1 = m(x − x1 )
V(1, 5)
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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 = −x
2. 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. 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 = −x
2. 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. 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 = −x
2. 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+ 4
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25. Example 1
3. Solve the system of these two equations.
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26. Example 1
3. Solve the system of these two equations.
⎧y = − x
⎨
⎩y = x + 4
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27. Example 1
3. Solve the system of these two equations.
⎧y = − x −x = x + 4
⎨
⎩y = x + 4
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28. Example 1
3. Solve the system of these two equations.
⎧y = − x −x = x + 4
⎨ −2x = 4
⎩y = x + 4
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29. Example 1
3. Solve the system of these two equations.
⎧y = − x −x = x + 4
⎨ −2x = 4
⎩y = x + 4
x = −2
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30. Example 1
3. Solve the system of these two equations.
⎧y = − x −x = x + 4 y = −(−2)
⎨ −2x = 4
⎩y = x + 4
x = −2
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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 = −2
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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 = −2
Thursday, January 5, 2012
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)
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34. Example 1
4. Use the distance formula utilizing this point on the line
and the point not on the line.
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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)
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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 2
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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 2
Thursday, January 5, 2012
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 2
Thursday, January 5, 2012
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 2
Thursday, January 5, 2012
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 2
Thursday, January 5, 2012
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 2
Thursday, January 5, 2012
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 2
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43. Example 2
Find the distance between the parallel lines m and n with
the following equations.
y = 2x + 3 y = 2x − 1
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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. 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
Thursday, January 5, 2012
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)
2
Thursday, January 5, 2012
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
2
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48. Example 2
2. Find the intersection of the perpendicular line and the
other parallel line using a system.
Thursday, January 5, 2012
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
⎩ 2
Thursday, January 5, 2012
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
2
Thursday, January 5, 2012
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
2
Thursday, January 5, 2012
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
2
Thursday, January 5, 2012
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
2
Thursday, January 5, 2012
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
2
Thursday, January 5, 2012
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
2
Thursday, January 5, 2012
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
2
Thursday, January 5, 2012
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
2
Thursday, January 5, 2012
58. Example 2
3. Use the new point and original y-intercept you chose in
step 2 in the distance formula.
Thursday, January 5, 2012
59. Example 2
3. 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. Example 2
3. 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 2
Thursday, January 5, 2012
61. Example 2
3. 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
Thursday, January 5, 2012
62. Example 2
3. 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 2
Thursday, January 5, 2012
63. Example 2
3. 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
Thursday, January 5, 2012
64. Example 2
3. 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
Thursday, January 5, 2012
65. Example 2
3. 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
Thursday, January 5, 2012
66. Example 2
3. 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 units
Thursday, January 5, 2012
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. 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. 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
Thursday, January 5, 2012
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 ≈ 2.83
Thursday, January 5, 2012
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 units
Thursday, January 5, 2012
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 Jefferson
Thursday, January 5, 2012
