Like this presentation? Why not share!

# AA Section 3-5

## by Jimbo Lamb, Math Teacher and Technology Coach at Annville-Cleona Secondary School on Nov 11, 2008

• 490 views

Finding an Equation of a Line

Finding an Equation of a Line

### Views

Total Views
490
Views on SlideShare
428
Embed Views
62

Likes
0
3
0

### 2 Embeds62

 http://mrlambmath.wikispaces.com 61 https://mrlambmath.wikispaces.com 1

## AA Section 3-5Presentation Transcript

• Section 3-5 Finding the Equation of a Line ...or th e Po in t-S lo pe Th eo re m se ct io n
• Warm-up Write an equation for the line through the pair of points a. (5, 9), (5, -2) b. (9, 1), (6, 4)
• Warm-up Write an equation for the line through the pair of points a. (5, 9), (5, -2) b. (9, 1), (6, 4) x=5
• Warm-up Write an equation for the line through the pair of points a. (5, 9), (5, -2) b. (9, 1), (6, 4) x=5 y = -x + 10
• Warm-up Write an equation for the line through the pair of points a. (5, 9), (5, -2) b. (9, 1), (6, 4) x=5 y = -x + 10 Not quite sure how this is done? We’ll see two ways today
• Question What determines a line?
• Question What determines a line? Two points
• Question What determines a line? Two points ...and once we have two points, we can find an equation
• Example 1 The formula relating blood pressure and age is linear. Normal systolic blood pressures are 110 for a 20 year old and 130 for a 60 year old. Graph the line and find an equation where blood pressure B is a function of age A.
• Example 1 The formula relating blood pressure and age is linear. Normal systolic blood pressures are 110 for a 20 year old and 130 for a 60 year old. Graph the line and find an equation where blood pressure B is a function of age A. 130.0 Blood Pressure 97.5 65.0 32.5 0 0 15 30 45 60 Age
• Example 1 (con’t) (20, 110), (60, 130)
• Example 1 (con’t) (20, 110), (60, 130) 130 − 110 m= 60 − 20
• Example 1 (con’t) (20, 110), (60, 130) 130 − 110 20 m= = 60 − 20 40
• Example 1 (con’t) (20, 110), (60, 130) 130 − 110 20 1 m= = = 60 − 20 40 2
• Example 1 (con’t) (20, 110), (60, 130) 130 − 110 20 1 m= = = 60 − 20 40 2 B= 2 A+b 1
• Example 1 (con’t) (20, 110), (60, 130) 130 − 110 20 1 m= = = 60 − 20 40 2 B= 2 A+b 1 110 = (20) + b 1 2
• Example 1 (con’t) (20, 110), (60, 130) 130 − 110 20 1 m= = = 60 − 20 40 2 B= 2 A+b 1 110 = (20) + b 1 2 110 = 10 + b
• Example 1 (con’t) (20, 110), (60, 130) 130 − 110 20 1 m= = = 60 − 20 40 2 B= 2 A+b 1 110 = (20) + b 1 2 110 = 10 + b b = 100
• Example 1 (con’t) (20, 110), (60, 130) 130 − 110 20 1 m= = = 60 − 20 40 2 B= 2 A+b 1 110 = (20) + b 1 2 110 = 10 + b b = 100 B = 2 A + 100 1
• Example 1 (con’t) (20, 110), (60, 130) 130 − 110 20 1 m= = = 60 − 20 40 2 B= 2 A+b 1 110 = (20) + b 1 2 110 = 10 + b b = 100 B = 2 A + 100 1 There has to be a better way!
• Point-Slope Theorem
• Point-Slope Theorem If a line contains (x1, y1) and has slope m, then it has the equation y - y1 = m(x - x1)
• Point-Slope Theorem If a line contains (x1, y1) and has slope m, then it has the equation y - y1 = m(x - x1) (In other words, you need a point and the slope)
• Example 2 Find an equation for the line through (-3, 6) and (5, 0) using the point-slope theorem.
• Example 2 Find an equation for the line through (-3, 6) and (5, 0) using the point-slope theorem. 6−0 m= −3 − 5
• Example 2 Find an equation for the line through (-3, 6) and (5, 0) using the point-slope theorem. 6−0 6 m= = −3 − 5 −8
• Example 2 Find an equation for the line through (-3, 6) and (5, 0) using the point-slope theorem. 6−0 6 3 m= = =− −3 − 5 −8 4
• Example 2 Find an equation for the line through (-3, 6) and (5, 0) using the point-slope theorem. 6−0 6 3 m= = =− −3 − 5 −8 4 y − y1 = m(x − x1 )
• Example 2 Find an equation for the line through (-3, 6) and (5, 0) using the point-slope theorem. 6−0 6 3 m= = =− −3 − 5 −8 4 y − y1 = m(x − x1 ) 3 y − 0 = − 4 (x − 5)
• Example 2 Find an equation for the line through (-3, 6) and (5, 0) using the point-slope theorem. 6−0 6 3 m= = =− −3 − 5 −8 4 y − y1 = m(x − x1 ) 3 y − 0 = − 4 (x − 5) 3 15 y=−4x+ 4
• When dealing with real world situations, deal with the problem as we always have: find the equation first, then answer the question.
• Example 3 The lightest recommended weight for a Martian with height 4’10” is 109 lbs. This weight increases 2 lbs/in to a height of 5’1” and then goes up 3 lbs/in to a height of 6’ which is tall for a Martian. a. Graph the situation
• Example 3 The lightest recommended weight for a Martian with height 4’10” is 109 lbs. This weight increases 2 lbs/in to a height of 5’1” and then goes up 3 lbs/in to a height of 6’ which is tall for a Martian. a. Graph the situation First, we need to convert all heights to inches
• Example 3 The lightest recommended weight for a Martian with height 4’10” is 109 lbs. This weight increases 2 lbs/in to a height of 5’1” and then goes up 3 lbs/in to a height of 6’ which is tall for a Martian. a. Graph the situation First, we need to convert all heights to inches 4’10” =
• Example 3 The lightest recommended weight for a Martian with height 4’10” is 109 lbs. This weight increases 2 lbs/in to a height of 5’1” and then goes up 3 lbs/in to a height of 6’ which is tall for a Martian. a. Graph the situation First, we need to convert all heights to inches 4’10” = 4(12) +10 =
• Example 3 The lightest recommended weight for a Martian with height 4’10” is 109 lbs. This weight increases 2 lbs/in to a height of 5’1” and then goes up 3 lbs/in to a height of 6’ which is tall for a Martian. a. Graph the situation First, we need to convert all heights to inches 4’10” = 4(12) +10 = 58”
• Example 3 The lightest recommended weight for a Martian with height 4’10” is 109 lbs. This weight increases 2 lbs/in to a height of 5’1” and then goes up 3 lbs/in to a height of 6’ which is tall for a Martian. a. Graph the situation First, we need to convert all heights to inches 4’10” = 4(12) +10 = 58” 5’1” =
• Example 3 The lightest recommended weight for a Martian with height 4’10” is 109 lbs. This weight increases 2 lbs/in to a height of 5’1” and then goes up 3 lbs/in to a height of 6’ which is tall for a Martian. a. Graph the situation First, we need to convert all heights to inches 4’10” = 4(12) +10 = 58” 5’1” = 5(12) + 1 =
• Example 3 The lightest recommended weight for a Martian with height 4’10” is 109 lbs. This weight increases 2 lbs/in to a height of 5’1” and then goes up 3 lbs/in to a height of 6’ which is tall for a Martian. a. Graph the situation First, we need to convert all heights to inches 4’10” = 4(12) +10 = 58” 5’1” = 5(12) + 1 = 61”
• Example 3 The lightest recommended weight for a Martian with height 4’10” is 109 lbs. This weight increases 2 lbs/in to a height of 5’1” and then goes up 3 lbs/in to a height of 6’ which is tall for a Martian. a. Graph the situation First, we need to convert all heights to inches 4’10” = 4(12) +10 = 58” 5’1” = 5(12) + 1 = 61” 6’ = 6(12) = 72”
• Example 3 The lightest recommended weight for a Martian with height 4’10” is 109 lbs. This weight increases 2 lbs/in to a height of 5’1” and then goes up 3 lbs/in to a height of 6’ which is tall for a Martian. a. Graph the situation First, we need to convert all heights to inches 4’10” = 4(12) +10 = 58” 5’1” = 5(12) + 1 = 61” 6’ = 6(12) = 72” (58”, 109 lbs)
• Example 3 The lightest recommended weight for a Martian with height 4’10” is 109 lbs. This weight increases 2 lbs/in to a height of 5’1” and then goes up 3 lbs/in to a height of 6’ which is tall for a Martian. a. Graph the situation First, we need to convert all heights to inches 4’10” = 4(12) +10 = 58” 5’1” = 5(12) + 1 = 61” 6’ = 6(12) = 72” (58”, 109 lbs) (61”, 115 lbs)
• Example 3 The lightest recommended weight for a Martian with height 4’10” is 109 lbs. This weight increases 2 lbs/in to a height of 5’1” and then goes up 3 lbs/in to a height of 6’ which is tall for a Martian. a. Graph the situation First, we need to convert all heights to inches 4’10” = 4(12) +10 = 58” 5’1” = 5(12) + 1 = 61” 6’ = 6(12) = 72” (58”, 109 lbs) (61”, 115 lbs) (72”, 148 lbs)
• Example 3 (con’t) (58”, 109 lbs) (61”, 115 lbs) (72”, 148 lbs) b. Find two equations that describe these situations
• Example 3 (con’t) (58”, 109 lbs) (61”, 115 lbs) (72”, 148 lbs) b. Find two equations that describe these situations y - y1 = m(x - x1)
• Example 3 (con’t) (58”, 109 lbs) (61”, 115 lbs) (72”, 148 lbs) b. Find two equations that describe these situations y - y1 = m(x - x1) w - w1 = m(h - h1)
• Example 3 (con’t) (58”, 109 lbs) (61”, 115 lbs) (72”, 148 lbs) b. Find two equations that describe these situations y - y1 = m(x - x1) w - w1 = m(h - h1) w - 109 = 2(h - 58)
• Example 3 (con’t) (58”, 109 lbs) (61”, 115 lbs) (72”, 148 lbs) b. Find two equations that describe these situations y - y1 = m(x - x1) w - w1 = m(h - h1) w - 109 = 2(h - 58) w = 2h - 7
• Example 3 (con’t) (58”, 109 lbs) (61”, 115 lbs) (72”, 148 lbs) b. Find two equations that describe these situations y - y1 = m(x - x1) w - w1 = m(h - h1) w - 109 = 2(h - 58) w = 2h - 7 for 58 ≤ h ≤ 61
• Example 3 (con’t) (58”, 109 lbs) (61”, 115 lbs) (72”, 148 lbs) b. Find two equations that describe these situations y - y1 = m(x - x1) w - w1 = m(h - h1) w - w1 = m(h - h1) w - 109 = 2(h - 58) w = 2h - 7 for 58 ≤ h ≤ 61
• Example 3 (con’t) (58”, 109 lbs) (61”, 115 lbs) (72”, 148 lbs) b. Find two equations that describe these situations y - y1 = m(x - x1) w - w1 = m(h - h1) w - w1 = m(h - h1) w - 109 = 2(h - 58) w - 115 = 3(h - 61) w = 2h - 7 for 58 ≤ h ≤ 61
• Example 3 (con’t) (58”, 109 lbs) (61”, 115 lbs) (72”, 148 lbs) b. Find two equations that describe these situations y - y1 = m(x - x1) w - w1 = m(h - h1) w - w1 = m(h - h1) w - 109 = 2(h - 58) w - 115 = 3(h - 61) w = 2h - 7 w = 3h - 68 for 58 ≤ h ≤ 61
• Example 3 (con’t) (58”, 109 lbs) (61”, 115 lbs) (72”, 148 lbs) b. Find two equations that describe these situations y - y1 = m(x - x1) w - w1 = m(h - h1) w - w1 = m(h - h1) w - 109 = 2(h - 58) w - 115 = 3(h - 61) w = 2h - 7 w = 3h - 68 for 58 ≤ h ≤ 61 for 61 < h ≤ 72
• Homework
• Homework p. 165 #1 - 21 “ I’m not sure I want popular opinion on my side -- I’ve noticed those with the most opinions often have the fewest facts.” - Bethania McKenstry