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Math Help - I can't troubleshoot this

  1. #1
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    I can't troubleshoot this

    I'm trying to solve for the coordinate that is 1/3 of the way between point (1,7) and point (10,4) USING the distance formula.


    This is what I had but I am stuck.




    Distance between point A and B:
    Distance = Sqrt ((Xsub2 - Xsub1)^2 + (Ysub2 - Ysub1)^2)

    Find the distance, multiply it by 1/3 - we'll call this distance ab.

    ab = sqrt ((Xsub2 - Xsub1)^2 + (Ysub2 - Ysub1)^2); Simplify and solve by pluging in the values.

    ab^2 = (Xsub2 - 1)^2 + (Ysub2 - 7)^2; where Xsub2 and Ysub2 are the coordinates you are looking for.

    Note: Keep in mind that point A is (1,7) and B is (10,4).
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  2. #2
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    Quote Originally Posted by Masterthief1324 View Post
    I'm trying to solve for the coordinate that is 1/3 of the way between point (1,7) and point (10,4) USING the distance formula.
    note the horizontal distance between the x-values ...

    x = 1 and x = 10 ... 9 units apart

    x=4 and x = 7 would be 1/3 and 2/3 of the way respectively.

    do the same in the vertical direction ...

    y = 7 and y = 4 ... 3 units apart

    y = 6 and y = 5 would be 1/3 and 2/3 of the way respectively

    so ,,,

    (1,7) (4,6) (7,5) (10,4)
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  3. #3
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    Quote Originally Posted by Masterthief1324 View Post
    I'm trying to solve for the coordinate that is 1/3 of the way between point (1,7) and point (10,4) USING the distance formula.


    This is what I had but I am stuck.




    Distance between point A and B:
    Distance = Sqrt ((Xsub2 - Xsub1)^2 + (Ysub2 - Ysub1)^2)

    Find the distance, multiply it by 1/3 - we'll call this distance ab.

    ab = sqrt ((Xsub2 - Xsub1)^2 + (Ysub2 - Ysub1)^2); Simplify and solve by pluging in the values.

    ab^2 = (Xsub2 - 1)^2 + (Ysub2 - 7)^2; where Xsub2 and Ysub2 are the coordinates you are looking for.

    Note: Keep in mind that point A is (1,7) and B is (10,4).
    D = \sqrt{(10 - 1)^2 + (4 - 7)^2}

     = \sqrt{9^2 + (-3)^2}

     = \sqrt{81 + 9}

     = \sqrt{90}

     = 3\sqrt{10}.


    Since there is some point, call it (x, y) which is between (1, 7) and (10, 4), which divides the distance into the ratio 1:2.

    That would mean that one length has magnitude \sqrt{10} and the other has magnitude 2\sqrt{10}.


    So \sqrt{10} = \sqrt{(x - 1)^2 + (y - 7)^2}

    10 = (x - 1)^2 + (y - 7)^2.


    Also 2\sqrt{10} = \sqrt{(10 - x)^2 + (4 - y)^2}

    40 = (10 - x)^2 + (4 - y)^2.


    Now you have two equations in two unknowns you can solve simultaneously.
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