Algebra: please check

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Find the equation of a line that passes through the point (1,2)
and has a slope of 1/3. Show how you found the equation.

Would the equation be:

y-2=1/3(x-1)

Do I need to finish this? I am not sure how.

  • Algebra: please check -

    You set it up right. Don't forget your plugging in for x and y and you need to finish this.

    So your equation would be

    -2=1/3(1)+b
    B would be your unknown y-intercept.
    But this is not in slope-intercept form
    So now you need to solve for b.

    -2=1/3(1)+b
    -2=1/3+b

    subtract 1/3 from both sides you get
    -2 1/3=b
    or
    -7/3=b


    your equation would then be
    y=1/3x-2 1/3
    or
    y=1/3x-7/3

  • Algebra: please check -

    There are several ways to write the equation of a straight line.
    The way troy showed you is the y-intercept-slope form.
    Your starting equation suggests that you have learned the point-slope form
    continue from
    y-2=1/3(x-1) by multiplying both sides by 3 to avoid fractions ...
    3y - 6 = x-1
    take everybody to the left side
    -x + 3y - 5 = 0
    usually we start with a positive x
    x - 3y + 5 = 0

    This is the general form of the equation of a straight line, and it contains no fractions.

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