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I Fnd the area A of the triangle with the given the vertices

  1. Nov 30, 2016 #1
    (0, 0), (3, 5), (1, 8)

    Find the slopes and equations for each line

    (0,0) ----> (3,5) = 5/3x
    (0,0)---->(1,8) = 8x
    (1,8)---->(3,5) = -3/2x+ 19

    Then I set up the integrals (on x)

    Integral sign from 0 to 1 (8x-5/3x)dx + Integral sign from 1 to 3 [(-3/2x+19)-5/3x) dx

    I got 117/4 as an answer and that's wrong. My algebra/arithmetic isn't wrong (i triple checked it) so it must have something to do with the set up. What's wrong with it?
     
    Last edited: Nov 30, 2016
  2. jcsd
  3. Nov 30, 2016 #2

    lewando

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    For the line specified by (1,8) and (3,5), what algebra/arithmetic did you use to get a y-intercept of 19?
     
  4. Nov 30, 2016 #3

    lurflurf

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    that should be

    (0,0) ----> (3,5) = 5/3x
    (0,0)---->(1,8) = 8x
    (1,8)---->(3,5) = -3/2x+ 19/2

    I suppose the directions asked for an integral it is not the best way otherwise
     
  5. Nov 30, 2016 #4

    lurflurf

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    3+8*2=19
    then divide by 2
     
  6. Dec 1, 2016 #5
    Wait, 19/2?

    y=mx+b
    For the slope i got (-3/2)
    used point (3,5)
    5=-3/2(3)+b
    10=-9
    19=b

    Did I do something wrong?
     
  7. Dec 1, 2016 #6

    lurflurf

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    should be
     
  8. Dec 1, 2016 #7
    Where did you get the 2b from?
     
  9. Dec 1, 2016 #8

    lurflurf

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    when you multiply both sides by 2 also multiply b
     
  10. Dec 1, 2016 #9

    Nidum

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    Have you been specifically told to use calculus for this problem ?

    If not then the area of the triangle can be calculated directly from the vertex coordinates .

    Try searching on ' finding area of a triangle using coordinate geometry '

    and for general interest ' Shoelace formula '

    If this is homework then it should really be in the PF homework section .
     
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