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Math Help - medians....hypotenuse.....yeah

  1. #1
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    medians....hypotenuse.....yeah

    Next topic is geometry....in class. I'm trying to get a head start before I'm left in the dust........again.

    Given the points A(-1,1), B(-2,-2) and C(4,0) as the vertices of a triangle ABC. Determine the length of the median AM:

    OR

    Given A(-3,0), B(3,-2) and C(1,-4)as the vertices of a triangle ABC. Show that the midpoint M of the hypotenuse is equidistant from the vertices of the triangle.

    BTW, I've learned how to find the midpoint of a line and length of a line........on my own ( _ ) so no need going over that....you know, if it has anything to do with understanding this.

    Thanks-a-Bunch!
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  2. #2
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    Quote Originally Posted by amai View Post
    Next topic is geometry....in class. I'm trying to get a head start before I'm left in the dust........again.

    Given the points A(-1,1), B(-2,-2) and C(4,0) as the vertices of a triangle ABC. Determine the length of the median AM:
    The mid point M of BC is ((-2+4)/2, (-2)/2) = (1, -1)

    The length of AM is: sqrt[(A(1)-M(1))^2 + (A(2)+M(2))^2]

    ..................................... = sqrt[(-1-1)^2 + (1+1)^2] = 2 sqrt(2)

    RonL
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  3. #3
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    Here is another way to do it (the nice thing about it is that you do not need coordinates to use it).

    Let ABC be a triangle with sides a,b,c.
    Let m_a represent the median drawn to side a, we have the following:

    m_a=(1/2)*sqrt(2(b^2+c^2)-a^2)

    Thus, you find all the three sides of this triagnle (through distance formula) and then you can apply that theorem.
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  4. #4
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    Hello, amai!

    Given: A(-3,0), B(3,-2) and C(1,-4)as the vertices of right triangle ABC.
    Show that the midpoint M of the hypotenuse is equidistant from the vertices of the triangle.

    BTW, I've learned how to find the midpoint of a line and length of a line. . Good!

    Now put all that knowledge together . . .

    Make a sketch and AB seems to be the hypotenuse.

    The midpoint of AB is: .x = ˝(-3 + 3) = 0, .y = ˝(0 -2) = -1
    . . Hence, we have: .M(0,-1)


    Now we're to show that: .MA = MB = MC.

    . . MA˛ .= .(-3 - 0)˛ + (0 + 1)˛ . . MA = √10

    . . MB˛ .= .(3 - 0)˛ + (-2 + 1)˛ . . MB = √10

    . . MC˛ .= .(1 - 0)˛ + (-4 + 1)˛ . . MC = √10

    . . . ta-DAA!

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