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Geometry and Discrete, prove this?

  1. Feb 12, 2006 #1
    The question asks:

    In triangle ABC, with vertices A(0,a), B(0,0), and C(b,c), prove that the right bisecors of the sides meet at a common point.

    Ok, this question is really getting to me. I know that I could find the equations of all three right bisectors and then solve for x and y through substitution. However, when I try to solve it through substitution I get 50 variables and no cancellations. I'm sure there is an easier way. Please help!
     
  2. jcsd
  3. Feb 12, 2006 #2

    AKG

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    No, you should be able to find the mid-point and slope of each side of the triangle. With this, you should be able to find the Cartesian equation of the right bisector of each side (note, the slope of the side AB is undefined, but you can still find the Cartesian equation for the right bisector; actually, the right bisector of this side has the easiest equation to find). This will give you 3 equations with 2 unknowns. You should be able to find a unique solution for these unknowns.
     
  4. Feb 12, 2006 #3
    I did just that. I found the midpoint of each side, took the slope, and then determined the equation of each right bisector. My one equation is y=a/2, whereas my other two eqns are huge. Is there a way to simplify the other two equations using y = a/2? Can you please show me how to prove this problem? I understand the idea, I just can't get it to work.
     
  5. Feb 12, 2006 #4

    AKG

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    No matter how "huge" your other two equations are, they must be of the form

    y = mx + d

    where m and d are just constants (in terms of a, b, and c, which are also just constants). You'll get three equations:

    y = a/2
    y = m1x + d1
    y = m2x + d2

    You want to prove that there is a unique solution to this problem. They must have taught you what it means to prove that there are exactly n solutions to a system of equations (in this case, n = 1), and how to do it. You can prove not only that there exists a unique solution to these equations, you can also say what that solution is.
     
  6. Feb 12, 2006 #5

    HallsofIvy

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    The segment from A(0,a) to B(0,0) has midpoint (0, a/2) is horizontal. The equation of the perpendicular bisector is, as you say, y= a/2.

    The segment from B(0, 0) to C(b, c) has midpoint (b/2, c/2) and slope c/b. The equation of the perpendicular bisector is
    y= -b/c(x- b/2)+ c/2.

    The segment from A(0,a) to C(b,c) has midpoint (b/2, (a+c)/2) and slope (c-a)/b. The equation of the perpendicular bisector is
    y= (b/(a-c))(x- b/2)+ (a+c)/2.

    They don't look all that "huge"! And I surely don't see where you got "50 variables". Even if you count a, b, c as variables (which they are not), that would still be only 5. Actually, there are only 2 "variables", x and y.
     
  7. Feb 12, 2006 #6
    I am so confused. I'm going to skip it and hope that a question like that isn't on my test. Thanks for your time.
     
  8. Feb 13, 2006 #7

    HallsofIvy

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    You know immediately that the y coordinate of the point of intersection must be a/2 because one of your lines always has that y coordinate. What happens if you set y= a/2 in the other two equations and solve for x?
     
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