• Share
  • Email
  • Embed
  • Like
  • Save
  • Private Content
Geometry Section 3-6 1112
 

Geometry Section 3-6 1112

on

  • 932 views

Perpendiculars and Dist

Perpendiculars and Dist

Statistics

Views

Total Views
932
Views on SlideShare
538
Embed Views
394

Actions

Likes
0
Downloads
0
Comments
0

1 Embed 394

http://mrlambmath.wikispaces.com 394

Accessibility

Categories

Upload Details

Uploaded via as Adobe PDF

Usage Rights

© All Rights Reserved

Report content

Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
  • Full Name Full Name Comment goes here.
    Are you sure you want to
    Your message goes here
    Processing…
Post Comment
Edit your comment

    Geometry Section 3-6 1112 Geometry Section 3-6 1112 Presentation Transcript

    • Section 3-6 Perpendiculars and DistanceThursday, January 5, 2012
    • 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
    • Vocabulary 1. Equidistant: 2. Distance Between a Point and a Line: 3. Distance Between Parallel Lines:Thursday, January 5, 2012
    • 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
    • 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
    • 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
    • Postulates & Theorems 1. Perpendicular Postulate: 2. Two Lines Equidistant from a Third:Thursday, January 5, 2012
    • 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
    • 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
    • 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
    • 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
    • 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
    • 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
    • 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
    • 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
    • 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
    • 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
    • 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
    • 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
    • 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
    • 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
    • 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
    • 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
    • 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
    • Example 1 3. Solve the system of these two equations.Thursday, January 5, 2012
    • Example 1 3. Solve the system of these two equations. ⎧y = − x ⎨ ⎩y = x + 4Thursday, January 5, 2012
    • Example 1 3. Solve the system of these two equations. ⎧y = − x −x = x + 4 ⎨ ⎩y = x + 4Thursday, January 5, 2012
    • Example 1 3. Solve the system of these two equations. ⎧y = − x −x = x + 4 ⎨ −2x = 4 ⎩y = x + 4Thursday, January 5, 2012
    • 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
    • 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
    • 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
    • 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
    • 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
    • Example 1 4. Use the distance formula utilizing this point on the line and the point not on the line.Thursday, January 5, 2012
    • 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
    • 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
    • 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
    • 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
    • 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
    • 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
    • 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
    • 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
    • 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
    • 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
    • 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
    • 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
    • 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
    • Example 2 2. Find the intersection of the perpendicular line and the other parallel line using a system.Thursday, January 5, 2012
    • 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
    • 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
    • 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
    • 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
    • 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
    • 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
    • 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
    • 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
    • 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
    • Example 23. Use the new point and original y-intercept you chose in step 2 in the distance formula.Thursday, January 5, 2012
    • 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
    • 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
    • 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
    • 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
    • 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
    • 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
    • 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
    • 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
    • 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
    • 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
    • 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
    • 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
    • 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
    • Check Your Understanding Review problems #1-8 on p. 218Thursday, January 5, 2012
    • Problem SetThursday, January 5, 2012
    • 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