This document discusses finding distances between lines and points. It defines equidistant lines as lines where the distance between them is the same when measured along a perpendicular. It explains that the distance between a point and line is the length of the perpendicular segment from the point to the line, and the distance between parallel lines is the length of the perpendicular segment between the lines. The document provides an example problem that finds the distance between a line and point by first finding the equations of the given line and perpendicular line through the point, then solving the system of equations.

1. Section 3-6
Perpendiculars and Distance
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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:
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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:
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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)
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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)
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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)
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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
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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
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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
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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
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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
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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
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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.
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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
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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
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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.
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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
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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
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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
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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
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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
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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
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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
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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
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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
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58. Example 2
3. Use the new point and original y-intercept you chose in
step 2 in the distance formula.
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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)
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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
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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
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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
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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
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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
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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
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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
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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).
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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:
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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
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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
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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
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72. Check Your Understanding
Review problems #1-8 on p. 218
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73. Problem Set
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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
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