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    A straight line graph goes through (0,8) and (6,0). Write its equation in the form ax+by=c, where a, b, and c are constants.

    Please give the method.
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    quick please
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    c is 8


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    (Original post by hello654321)
    A straight line graph goes through (0,8) and (6,0). Write its equation in the form ax+by=c, where a, b, and c are constants.

    Please give the method.
    1) find gradient:
     m = \frac{y1-y2}{x1-x2}

    2) Find equation and rearrange:
     y - y1 = m(x-x1)
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    (Original post by asinghj)
    1) find gradient:
     m = \frac{y1-y2}{x1-x2}

    2) Find equation and rearrange:
     y - y1 = m(x-x1)
    0,8 6,0
    That isn't the right method i don't think...
    gradient is 0-8/6-0 = -8/6

    y - 8 = -8/6(x-0)
    y-8= -8/6x - 0
    +8
    y= -8/6 x + 8 ???

    this isn't the form ax+by=c and also i checked the answer and its 4x+3y=24??
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    (Original post by hello654321)
    0,8 6,0
    That isn't the right method i don't think...
    gradient is 0-8/6-0 = -8/6

    y - 8 = -8/6(x-0)
    y-8= -8/6x - 0
    +8
    y= -8/6 x + 8 ???

    this isn't the form ax+by=c and also i checked the answer and its 4x+3y=24??
     - \frac{8}{6} simplifies to  - \frac{4}{3}. Then multiply the whole expression by the denominator of the fraction, rearrange into the required form, and you get the answer.
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    (Original post by K-Man_PhysCheM)
     - \frac{8}{6} simplifies to  - \frac{4}{3}. Then multiply the whole expression by the denominator of the fraction, rearrange into the required form, and you get the answer.
    ahhh i see thanks!
 
 
 
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