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Math Help - College Algebra Distance Formula and Mid point

  1. #1
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    College Algebra Distance Formula and Mid point

    Here I have two problems each for Distance Formula and Midpoint Formula

    I have solved these problems but would like to know if I am correct in my solving of the problems.

    Distance Forumla

    Give exact answer and where appropriate, an approximation to three decimal places.

    1) (4,6) and (5,9)

    d= √(5-4) + (9-6) =
    √(1) + (3) =
    √1+9 =
    √10 =
    3.162

    2) (-11, -8) and (1, -13)
    d= √(1-(-11) + (-13 -(-8) =
    √(1+11)+ (-13+8) =
    √(12) + (-5) =
    √144 + 25 =
    √169 =
    13

    Mid Point Formula

    1) (4, -9) and (-12, -3) =

    (4+ -12)/2, (-12 + -3)/2 =
    (-8)/2, (-15)/2 =
    (-4, -7.5)

    I didn't know if I should have left the answer for (-12 + -3)/2 as a fraction instead of actually dividing it and getting a decimal.

    2) (0, 1/2), (-2/5, 0) =
    (0+1/2)/2, (-2/5 +0)/2 =
    (1/2, -2/5) this is supposed to be a fraction and not divided by 2 at this point of the problem. This is the answer I got.


    Please let me know where I might have went astray or if I do well on these particular problems. Any help would be appreciated as I don't really know what I am doing.

    Thanks Much,
    Tricey36
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  2. #2
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    You've done fine on the distance formula, but the midpoint is actually:

    M = \left( \frac{x_1 + x_2}{2}, \frac{y_1+y_2}{2}\right)

    Not:
    \left( \frac{x_1 + y_1}{2}, \frac{x_2+y_2}{2}\right)
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  3. #3
    Moo
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    Hello !

    The part about distance is ok
    There's a problem with the midpoints ^^'

    Mid Point Formula

    1) (4, -9) and (-12, -3) =

    (4+ -12)/2, (-12 + -3)/2 =
    (-8)/2, (-15)/2 =
    (-4, -7.5)

    I didn't know if I should have left the answer for (-12 + -3)/2 as a fraction instead of actually dividing it and getting a decimal.

    2) (0, 1/2), (-2/5, 0) =
    (0+1/2)/2, (-2/5 +0)/2 =
    (1/2, -2/5) this is supposed to be a fraction and not divided by 2 at this point of the problem. This is the answer I got.
    For points M(x,y) and N(x',y'), the midpoint P has coordinates :

    P\left(\tfrac{x+x'}{2}~,~\tfrac{y+y'}{2}\right)
    In the first question, you did -12+(-3), which was actually -9+(-3)

    For the second one, you added up the coordinates of a same point oO or you inverted absciss and ordinate.

    It should've been (0+(-2/5))/2, (1/2+0)/2

    Note that \frac{\tfrac ab}{c}=\frac{a}{bc}

    As to know if it's important to put in fraction or in approximate value, it's up to you. It's always better to keep the exact values in order to have them within your reach if you need them again
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  4. #4
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    You did so well on the Distance Formula. What happened on the Midpoints? It looks like you just lost track of where you were, which, again, is very odd since you did it perfectly on the Distance Formula.

    MidPoint

    On the first, you used the 12 twice and never got to the 9. Try that again.

    On the second, you used the coordiantes from the same point, rather than the corresponding x- and y-coordinates from the two points. Try that again.
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  5. #5
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    Thanks

    Thanks to all who responded so quickly to my post about the distance and midpoint formula problems.
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