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Math Help - find the area coordinate triangle

  1. #1
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    find the area coordinate triangle

    Hello

    Given 3 points:
    (3,2) (-3,-3) (5,-1) are the vertices of a triangle. Find the area of the triangle.
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  2. #2
    Eater of Worlds
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    Do you have to use a particular method?.

    Try using the distance formula to find the lengths of the sides, then use Heron's formula.

    You could show off and find the equations of the lines that make up the triangles sides and then integrate.

    \int_{-3}^{3}\left[(\frac{5x}{6}-\frac{1}{2})-(\frac{x}{4}-\frac{9}{4})\right]dx+\int_{3}^{5}\left[(\frac{-3x}{2}+\frac{13}{2})-(\frac{x}{4}-\frac{9}{4})\right]dx
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  3. #3
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    i tried using ABS formula
    Let (x1,y1)=(3,2) ; (x2,y2)=(-3,-3) ; (x3-y3) = (5,-1)
    ABS(x2*y1-x1*y2+x3*y2-x2*y3+x1*y3-x3*y1)/2
    =ABS[(-3)(2)-(3)(-3)+(5)(-3)-(-3)(-1)+(3)(-1)-(5)(2)]/2
    =ABS(-6+9-15-3-3-10)/2
    =ABS(-28)/2
    =-14

    for some reason i get -14...am i suppose to get a "-" for an area??
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by subzero06 View Post
    i tried using ABS formula
    Let (x1,y1)=(3,2) ; (x2,y2)=(-3,-3) ; (x3-y3) = (5,-1)
    ABS(x2*y1-x1*y2+x3*y2-x2*y3+x1*y3-x3*y1)/2
    =ABS[(-3)(2)-(3)(-3)+(5)(-3)-(-3)(-1)+(3)(-1)-(5)(2)]/2
    =ABS(-6+9-15-3-3-10)/2
    =ABS(-28)/2
    =-14

    for some reason i get -14...am i suppose to get a "-" for an area??
    i didn't check your calculations, but if you take the absolute value of a negative number, the answer is positive. so ABS(-28)/2 = 28/2 = 14

    so provided your other calculations are correct, you would get +14 as the answer, which makes more sense than a negative answer
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  5. #5
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    oooh yea i get it now
    thank you very much!
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