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Thread: Finding a point on a line

  1. #1
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    Finding a point on a line

    So the formula for finding it has been explained to me as thus:

    BD/DA=CE/EA

    I understand the basis for this formula, what I struggle with is the item identification. So in the case of A=(-10,-4), and C= (8,5), find the y coordinate for B when x=-1. The example I'm given is as so

    BD/(-1-(-10))=(5-(-4))/(8-(-10)), or more cleanly BD/9=9/18

    It then breaks down to BD=9/2, which I understand. It then tells me since the B is higher than D (which makes 0 sense to me), that B is the y coord of D plus the length of BD, therefore -4+(9/2) (or 1/2).

    Can someone help explain the item identification and the last part of why B is D+BD?
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    Re: Finding a point on a line

    Quote Originally Posted by Tyro View Post
    So the formula for finding it has been explained to me as thus:

    BD/DA=CE/EA

    I understand the basis for this formula, what I struggle with is the item identification. So in the case of A=(-10,-4), and C= (8,5), find the y coordinate for B when x=-1. The example I'm given is as so

    BD/(-1-(-10))=(5-(-4))/(8-(-10)), or more cleanly BD/9=9/18

    It then breaks down to BD=9/2, which I understand. It then tells me since the B is higher than D (which makes 0 sense to me), that B is the y coord of D plus the length of BD, therefore -4+(9/2) (or 1/2).

    Can someone help explain the item identification and the last part of why B is D+BD?
    Please state the question exactly as given to you. Please do not try to give your interpretation. Frankly what you posted is meaningless.
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    Re: Finding a point on a line

    You'll need to provide more context about this "formula" ... frankly, I don't know what you're saying.

    For your problem, I would do it as follows ...

    The equation for line AC in slope-intercept form {recall ... $y-y_1 = m(x-x_1)$} is

    $y - 5 = \dfrac{5-(-4)}{8-(-10)}(x - 8)$

    $y - 5 = \dfrac{1}{2}(x - 8)$

    $x = -1 \implies y-5 = -\dfrac{9}{2} \implies y = \dfrac{1}{2}$
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    Re: Finding a point on a line

    The questions was stated exactly as it was given to me. The point A (-10,-4), and the point C (8,5) are 2 points on a line. Find the y coordinate for B (-1,y) (B is also on the same line)
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    Re: Finding a point on a line

    Quote Originally Posted by Tyro View Post
    The questions was stated exactly as it was given to me. The point A (-10,-4), and the point C (8,5) are 2 points on a line. Find the y coordinate for B (-1,y) (B is also on the same line)
    Well, the solution is given in my previous post ... only "formula" necessary is the point-slope form for finding the linear equation between two given points on the plane.
    Thanks from Tyro
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    Re: Finding a point on a line

    So to apply your formula to another equation so I can make sure I understand,

    A=(-7, -10) C=(10,-1), B= (2, y)

    I'd go y-(-1)=(9/17)(-8)
    y+1=-72/17
    y=-89/17?
    Last edited by Tyro; Jul 27th 2017 at 10:44 AM.
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    Re: Finding a point on a line

    Quote Originally Posted by Tyro View Post
    So to apply your formula to another equation so I can make sure I understand,

    A=(-7, -10) C=(10,-1), B= (2, y)

    I'd go y-(-1)=(9/17)(-8)
    y+1=-72/17
    y=-89/17?
    that will work ...
    Thanks from Tyro
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    Re: Finding a point on a line

    Quote Originally Posted by Tyro
    The questions was stated exactly as it was given to me. The point A (-10,-4), and the point C (8,5) are 2 points on a line. Find the y coordinate for B (-1,y) (B is also on the same line)
    Quote Originally Posted by skeeter View Post
    Well, the solution is given in my previous post ... only "formula" necessary is the point-slope form for finding the linear equation between two given points on the plane.
    It is sufficient to use the slope formula. Line segments AB, BC, and AC all lie on the same line and have the same slope. Pick any pair of slopes among the line segments AB, BC, and AC.

    Calculate the two slopes and set them equal to each other. You'll be guaranteed that there will be at least one y-variable in the equation. Solve for y.

    I have decided to work the slopes of line segment AB and line segment AC:


    \dfrac{y - (-4)}{-1 - (-10)} \ = \ \dfrac{5 - (-4)}{8 - (-10)}

    \dfrac{y  + 4}{-1  + 10} \ = \ \dfrac{5 + 4}{8 + 10}

    \dfrac{y + 4}{9} \ = \ \dfrac{9}{18}

    \dfrac{y + 4}{9} \ = \ \dfrac{1}{2}

    2(y + 4) \ = \ 9(1)

    2y + 8  \ = \  9

    2y \ = \ 1


    y \ = \ \dfrac{1}{2}
    Last edited by greg1313; Jul 27th 2017 at 02:24 PM.
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