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Distance between points

  1. Apr 26, 2010 #1
    1. The problem statement, all variables and given/known data

    A triangle is defined by the following vertices: (40,85), (15,0), and (-40,35). Is it a right triangle?

    2. Relevant equations

    Distance Formula:

    Given two points A(1,2) and B(3,4) Find the distance using the below forumla.

    d(A,B) = sqrt((x2-x1)^2 + (y2-y1)^2)


    3. The attempt at a solution

    1) We have X(40,85), Y(15,0), and Z(-40,35)

    2) We can use the distance formula b/c it is derived from pythagorean therom (which determines if a triangle is a right triangle).

    3) Using the distance formula for three vertices given two components for each might look something like this:

    d(x,y,z) = sqrt((x3-x2-x1)^2 + (y3-y2-y1)^2)

    ~107?
     
  2. jcsd
  3. Apr 26, 2010 #2
    Distances are between 2 points. You want to find the 3 distances between the 3
    possible pairs of points, and then check wether the square of one of them is equal
    to the sum of squares of the other 2.

    Another option is to draw vectors between the points and check wether the inner
    product of a pair of them is equal to 0.
     
  4. Apr 26, 2010 #3
    Okay, basically using my three points as a, b, and c, where, a(40,85), b(15,0), and c(-40,35).

    using c^2 = a^2 + b^2
    [tex]\Delta[/tex]c^2 = [tex]\Delta[/tex]a^2 + [tex]\Delta[/tex]b^2 ??
     
  5. Apr 26, 2010 #4
    1. The problem statement, all variables and given/known data

    A triangle is defined by the following vertices: (40,85), (15,0), and (-40,35). Is it a right triangle?

    Just wondering if what i have here is correct:

    d(x,y) = sqrt((x2-x1)^2+(y2-y1)^2)
    = sqrt((85-40)^2+(0-15)^2)
    = sqrt((45)^2+(-15)^2)
    = sqrt(2025+225)
    = 47.43

    d(x,z) = sqrt((x2-x1)^2+(z2-z1)^2)
    = sqrt(2025+(35-(-40))^2)
    = sqrt(2025+(75)^2)
    = sqrt(2025+5625)
    = 87.5

    d(y,z) = sqrt((y2-y1)^2+(z2-z1)^2)
    = sqrt((0-15)^2+5625)
    = sqrt((-15)^2 +5625)
    = sqrt(225+5625)
    = 76.5

    We found the longest side and we have now a,b,c values to use for pythagoream theroem:

    d(x,z)^2 = d(y,z)^2 + d(x,y)^2

    87.5^2 = 76.5^2 + 47.43^2

    7656.25 = 5852.25 + 2249.6

    7656.25 != 8101.85


    So it is not a rightangle.
     
  6. Apr 27, 2010 #5
    your notation d(x,y) makes no sense.

    The distance between (x1,y1) and (x2,y2) is sqrt ((x2 - x1)^2 + (y2-y1)^2)

    you need to subtract the x and the y coordinates of two different points, not the
    x and y coordinates of the same point.
     
  7. Apr 27, 2010 #6
    Alright, in continuation to the original problem:

    A shape is defined by the following vertices: (40,85), (15,0), and (-40,35) Is it a right triangle?

    I think that each point is a vector for a side of the potential right triangle.


    We can use the following vector formula:

    V1 = ((x1-x2)-(y1-y2))
    V2 = ((x1-x3)-(y1-y3))
    V3 = ((x2-x3)-(y2-y3))

    V1 = (40-15,85-0) = (25.85) Substitute our values and find the coordinate for a side
    V2 = (40-(-40), 85-35) = (80,50)
    V3 = (15-(-40), 0-35) = (55,-35)

    We now use the following matrix formula:


    A
    (C D) = AC + BD;
    B





    V1^t * V2 = 25 (80,50) = 25*80 + 85 * 50
    85 = 6250





    V1^t * V3 = 25 (55,-35) = 25*55 + 85 * -35
    85 = -1600





    V2^t * V3 = 80 (55,-35) = 80*55 + 50 * -35
    50 = 2650



    V1^t * V2 = 6250
    V1^t * V3 = -1600
    V2^t * V3 = 2650
    Based on the multiplications for each vector and matrix this is not a right angle, but an orthogonal.
     
  8. Apr 28, 2010 #7
    This is correct. Only the angles are not right angles because the inner product (or dot products) are not equal to 0. If two vectors are orthogonal, it means they are perpendicular BTW.
     
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