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Creating an equation that models a scenario

  1. Apr 20, 2016 #1
    1. The problem statement, all variables and given/known data
    5.The small town of Pi-ville wants to construct a road that connects to the Linear Super Highway. The road must provide a route that covers the shortest distance possible from the town to the highway. Write an equation that models the best route for the road.
    upload_2016-4-20_13-22-5.png



    2. Relevant equations
    y=mx+b


    3. The attempt at a solution
    ok so im guessing that i have to draw a line from Pi-ville to the linear super highway. Then determine the slope and y-intercept of that line to develop an equation that represent the best route for the road.
    i chose to draw a line from pi-ville to the point (0,4) on the linear highway.
    upload_2016-4-20_13-31-2.png
    then i calculated the slope of the line and determined the y-intercept

    (-4, 0) (0, 4)
    slope = (y2-y1) /(x2-x1)
    slope = (4-0) / 0-(-4)
    slope = 4/4 = 1

    y-intercept: +4 (since the line crosses the vertical axis at point 4)

    an equation to represent the best route for the road is:
    y=mx + b
    y = 4/4x + 4

    is this right?
     
  2. jcsd
  3. Apr 20, 2016 #2

    BvU

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    My guess reading is that here 'best' means 'shortest'. What is the characteristic of the shortest path to the blue line ?
     
  4. Apr 20, 2016 #3
    do you mean that there is a path from pi-ville to the linear highway that is shorter than the path i chose?
     
  5. Apr 20, 2016 #4

    PeroK

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    He does mean precisely that!
     
  6. Apr 20, 2016 #5
    upload_2016-4-20_14-9-51.png

    is this a shorter route?
     
  7. Apr 20, 2016 #6

    PeroK

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    That looks better. You just need the equation now.
     
  8. Apr 20, 2016 #7
    (-4, 0) (-1,4.8)
    slope = (y2-y1) /(x2-x1)
    slope = (4.8-0) / (-1 -(-4)) = 4.8/3

    y-intercept: +6 (since the line crosses the vertical axis at point 6)

    an equation to represent the best route for the road is:
    y=mx + b
    y = 4.8/3x + 6

    right?
     
  9. Apr 20, 2016 #8

    PeroK

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    You need to write ##(4.8/3)x##

    What happens when ##y = 0##? Where is the x-intercept from your equation?
     
  10. Apr 20, 2016 #9
    i got -3.75 after substituting 0 in for y in my equation?
     
  11. Apr 20, 2016 #10

    PeroK

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    Yes, but it should be ##-4##.
     
  12. Apr 20, 2016 #11
    so what does -4 represent?
     
  13. Apr 20, 2016 #12

    PeroK

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    The x-intercept you can see on your diagram!
     
  14. Apr 20, 2016 #13
    ooh i get it, so does this mean my answer is correct, or is it a few decimal places off, since i got -3.75 instead of -4
     
  15. Apr 20, 2016 #14

    PeroK

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    I think you can safely say that your equation is not correct!
     
  16. Apr 20, 2016 #15
    ok, so instead of using the point (-1, 4.8), i used (0, 6)
    and i got the equation y=1.5x + 6
    i get -4 as the x-intercept when substituting 0 in for y
    im assuming this is correct?
     
  17. Apr 20, 2016 #16

    PeroK

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    That looks better.
     
  18. Apr 20, 2016 #17
    thanks for the help :)
     
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