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Vertices of triangle, multi-problems

  1. Apr 30, 2008 #1
    Vertices of triangle ABC are A(2a,0), B(2b,0), C(0,2).

    a) Find the equations of the sides (check, did that)

    AB: y=0
    AC: x+ay=2a
    BC: x+by=2b

    I'm having trouble with b)

    Show that the equations of the medians are: x+(2a-b)y=2a, x+(2b-a)y=2b, 2x+(a+b)y=2(a+b)

    Ok, they're not referring to the midpoints of AB, AC, and BC? I think that's where my mistake is. The median is point of intersection where a line intersects each line at each line's midpoint and meets at a common center point inside the triangle?
  2. jcsd
  3. Apr 30, 2008 #2


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    Part of your objective is to find the midpoint of each side because that is an essential component of Median of a Triangle. The point in the interior of the triangle at which the medians intersect is of no concern in the answer.
  4. Apr 30, 2008 #3


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    No. A "median" of a triangle is the line segment from one vertex to the midpoint of the opposite side. For example, the midpoint of AB is (a+b, 0) and C is (0,2). What is the equation of that line?
  5. Apr 30, 2008 #4
    Your answer to part a) ...

    .... isn't quite right. Hint: What is the y-int of lines connecting A and C ? A and B?
  6. Apr 30, 2008 #5
    Sorry, I meant B and C?
  7. Apr 30, 2008 #6


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    The equations given were AC: x+ay=2a and BC: x+by=2b with A= (2a,0), B= (2b, 0), and C= (0, 2). When x= 2a, the first equation gives 2a+ ay= 2a or y= 0 and when x= 0 it gives ay= 2a or y= 2. When x= 2b, the second equation gives 2b+ by= 2b or y= 0 and when x= 0 it gives by= 2b so y= 2. Exactly what is wanted.
  8. May 1, 2008 #7

    Yup, yup and yup. Sheesh, the careless algebra mistakes I can make are embarrassing. thanks H.O.V.
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