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Thread: Standard Form Problem

  1. #1

    Standard Form Problem

    I'm having some trouble with this problem in my test review. Here it is:

    Find the equation of each line in standard form (where A > or = 0, and A, B, and C are integers):
    Slope is (c/d), contains (a, b)

    This was in a group of problems, and I didn't have any trouble with the others (they all had numbers instead of variables). To start this one, I put it into point-slope form like I had with the others.
    So I had: y - b = (c/d) * (x - a)
    y - b = (cx/d) - (ac/d)
    y = (cx/d) - (ac/d) + b

    I'm at a loss as to how I can put this (if it's even correct so far ) into the standard form. I checked the key and the final answer should be cx - dy = ac - bc. Any help on how to get there would be much appreciated! Thanks. Let me know if anything I wrote was unclear

  2. #2
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    Quote Originally Posted by justjess View Post
    I'm having some trouble with this problem in my test review. Here it is:

    Find the equation of each line in standard form (where A > or = 0, and A, B, and C are integers):
    Slope is (c/d), contains (a, b)

    This was in a group of problems, and I didn't have any trouble with the others (they all had numbers instead of variables). To start this one, I put it into point-slope form like I had with the others.
    So I had: y - b = (c/d) * (x - a)
    y - b = (cx/d) - (ac/d)
    y = (cx/d) - (ac/d) + b
    ....Next step: Multiply through by d to clear fractions
    I'm at a loss as to how I can put this (if it's even correct so far ) into the standard form. I checked the key and the final answer should be cx - dy = ac - bc. Any help on how to get there would be much appreciated! Thanks. Let me know if anything I wrote was unclear
    Using upper-case letters for the standard form, the line is
    [tex]A\ x +B\ y = C[/tex]

    The given line in point-slope form is as you said:
    [tex] y - b = \dfrac cd \times (x - a)[/tex]
    Multiply through by [tex]d[/tex] to clear fractions:
    [tex]d(y - b) = c(x-a) [/tex]
    [tex]dy - db = cx - ca[/tex]
    Transpose x and y terms to left, constants to right
    [tex]-cx + dy = -ca +db[/tex]
    Change signs so coefficient of x is positive
    [tex]cx - dy = ca - db[/tex]

    Equate coefficients of the standard form:
    [tex]A = c[/tex],.....[tex]B = -d[/tex],.....[tex]C = ca - db[/tex]
    DrPhil (not the TV guy)
    If we knew what we were doing,we wouldn't have to do it

  3. #3
    Elite Member
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    Quote Originally Posted by justjess View Post
    y = (cx/d) - (ac/d) + b
    Ok dokay Jess; what do you get if you multiply above equation by d ?
    I'm just an imagination of your figment !

  4. #4
    Quote Originally Posted by Denis View Post
    Ok dokay Jess; what do you get if you multiply above equation by d ?
    I did:
    y = (cx/d) - (ac/d) + b
    yd = cx - ac + bd
    yd - bd = cx - ac
    -cx + yd = -ac + bd
    cx - yd = ac - bd
    cx - dy = ac - bd
    (I just swapped the yd for dy here to make it look better)

    I just checked mine with yours, DrPhil, and it looks the same, but somehow the teacher got cx - dy = ac - bc. Well, the key has been wrong before! What's more important is I know how to do it now. Thanks all!

  5. #5
    Elite Member
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    Quote Originally Posted by justjess View Post
    cx - dy = ac - bd (I just swapped the yd for dy here to make it look better)

    ...but somehow the teacher got cx - dy = ac - bc.
    Teacher evidently made a goof...
    I'm just an imagination of your figment !

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