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Math Help - trisect lines

  1. #1
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    trisect lines

    - points x and y lie on cordinate axises
    - the line CD is disected into 3 parts by x and y equally (trisected is the term?)
    - we are given the value of C to find the possible cordinates of D

    Can someone lead me in the right direction, I'm not exactly looking for answers here.

    (this is for saturday school math, no idea how to, ty, im in grade 9 btw)
    Last edited by girlsrox32; October 4th 2008 at 07:18 PM.
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  2. #2
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    Quote Originally Posted by girlsrox32 View Post
    - points x and y lie on cordinate axises
    - the line CD is disected into 3 parts by x and y equally (trisected is the term?)
    - we are given the value of x to find the possible cordinates of y

    Can someone lead me in the right direction, I'm not exactly looking for answers here.

    (this is for saturday school math, no idea how to, ty, im in grade 9 btw)
    what do we know about CD? in particular, it's end points? how about using the distance formula? the distance between x and the end point close to x would be equal to the distance between y and the end point close to y. you would be able to set up an equation with y as the only unknown
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  3. #3
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    Sorry, edit of the first post! didnt see something
    its suppose to be
    - we are given the value of C to find the possible cordinates of D

    T_T can u rehelp me?
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  4. #4
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    Quote Originally Posted by girlsrox32 View Post
    uh we don't know one of CD's endpoints. but we know that x and y lie in coordinate axises.
    Alright then, say you know the location of C and the location of X. Then either Y is the midpoint of CX or X is the midpoint of CY.

    For example, say C = (1,3) and X = (7, 9).

    If Y is the midpoint of CX, then Y = (4, 6). and D = X + <CY> = (7, 9) + <3, 3> = (10, 12).

    If X is the midpoint of CY, then Y = (13, 15) and D = Y + <CX> = (13, 15) + <6, 6> = (19, 21).
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    Quote Originally Posted by icemanfan View Post
    Alright then, say you know the location of C and the location of X. Then either Y is the midpoint of CX or X is the midpoint of CY.

    For example, say C = (1,3) and X = (7, 9).

    If Y is the midpoint of CX, then Y = (4, 6). and D = X + <CY> = (7, 9) + <3, 3> = (10, 12).

    If X is the midpoint of CY, then Y = (13, 15) and D = Y + <CX> = (13, 15) + <6, 6> = (19, 21).
    But if x and y lie on the coordinate axises, wouldn't we have a value and variable as a ordered pair for each one? and there's two axises, so wouldn't there be two different pairs?
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  6. #6
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    Ok, I have a better understanding of the question now. You are given C only, and you have to find D, but you know that X and Y lie on the coordinate axes. Assuming that C does not lie on an axis, then you will have two possibilities for the location of D. Let C = (a, b) and define X to be the midpoint of CY. Then, because Y lies on an axis, it is either (0, y) for some y or (x, 0) for some x. If Y is (0, y), then X = \left(\frac{1}{2}a, 0\right), which means Y = (0, -b). If Y is (x, 0), then X = \left(0, \frac{1}{2}b\right), which means Y = (-a, 0). Can you determine the possible locations of D from this information?
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  7. #7
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    ^thanks!

    could you explain how you got to[ with a formula or something?] (1/2a, 0). I don't get why there would be 1/2
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  8. #8
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    X is the midpoint of CY. So if Y = (0, y) and C = (a, b), then the x-coordinate of X must be the average of the x-coordinates of C and Y, and since this value is nonzero and X is on an axis, the y-coordinate of X must be 0.
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  9. #9
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    omg ty!

    and uh, how did you conclude to Y(0,-b) after finding the x-coordinate midpoint. because when I do it, I have two variables to average out. and i also know that once i have the two coordinates of X and Y, I can find the distance between X and Y using that distance to substitute in a distance formula of Y and D, but for D I have two variables, how would I solve for each.
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  10. #10
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    Quote Originally Posted by girlsrox32 View Post
    and uh, how did you conclude to Y(0,-b) after finding the x-coordinate midpoint. because when I do it, I have two variables to average out. and i also know that once i have the two coordinates of X and Y, I can find the distance between X and Y using that distance to substitute in a distance formula of Y and D, but for D I have two variables, how would I solve for each.
    The average of the y-coordinates of C and Y must be zero. So you have \frac{b + y}{2} = 0 , hence

    b + y = 0
    y = -b

    Also, if you are given C = (a, b), then a and b are not variables. They are constants.
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  11. #11
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    but for D, i have only variables o_O
    , right?
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  12. #12
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    Nope. The same vector that connects X and Y also connects Y and D, and you know exactly what X and Y are (in each of the two cases), so you can find that vector and then add it to Y to obtain D.
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  13. #13
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    uh sry T_T but whats a vector
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  14. #14
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    A vector is a quantity that has length and direction. For example, the vector connecting the two points (1,1) and (4,5) is <3, 4> and it has a length of 5. It represents, in this case, the difference between the coordinates of the two points. If I add <3, 4> to (4, 5) I get (7, 9). Because I added the same vector twice, it turns out that (4, 5) is the midpoint of the line segment connecting (1, 1) and (7, 9).
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