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Math Help - Question about standard form of a line

  1. #1
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    Question about standard form of a line

    Hi, I'm new here! I'm having an issue with this problem in my homework. I need to find the equation of each line and put it in standard form.

    The points I have are (-7, -4) and (0,6).

    So I used the y2-y1 method.

    (6-4)
    -----
    (0-(-7) and got 10/7. So y= 10/7x + b
    . Then I found b.

    -4 = 10/7(-7) + b
    -4 = 10 + b
    -6 = b
    so in slope-intercept form my answer is y= 10/7x - 6.

    But i need it in standard form and the tutorials I've found for this say you just need to minus mx from both sides. So I did and I got -10/7x +y = -6. However my book says the answer should be 2x + 7y = -42.

    I think I'm not understanding how to convert slope-intercept form to standard form. The other problems in my homework have the same issue.

    By the way, I hope I put this in the correct format- sorry if I didn't! Thank you...
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  2. #2
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    The standard form has integer coefficients.
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  3. #3
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    Quote Originally Posted by Nycea View Post

    The points I have are (-7, -4) and (0,6).

    So I used the y2-y1 method.

    (6-4)
    -----
    (0-(-7) and got 10/7. So y= 10/7x + b
    . Then I found b.
    [I][B]
    -4 = 10/7(-7) + b
    next step is \displaystyle -4=-10+b \implies b = 6\neq -6
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  4. #4
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    Quote Originally Posted by Nycea View Post
    The points I have are (-7, -4) and (0,6).
    (6-4)
    -----
    (0-(-7) and got 10/7. So y= 10/7x + b
    . Then I found b.

    -4 = 10/7(-7) + b
    -4 = 10 + b
    -6 = b
    so in slope-intercept form my answer is y= 10/7x - 6.
    Yout slope is correct.
    Your b is wrong: should be 6; 10/7(-7) = -10
    Also quite evident from the given point (0,6)
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  5. #5
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    I think I've got it! Thanks you guys! <3
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