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how do you find the equation line of these ordered pairs (1/6, -1/3) and (5/6, 5)?

  • math -

    A common form of a linear equation in the two variables x and y is:

    y=mx+b

    where m and b designate constants. The origin of the name "linear" comes from the fact that the set of solutions of such an equation forms a straight line in the plane. In this particular equation, the constant m determines the slope or gradient of that line, and the constant term "b" determines the point at which the line crosses the y-axis, otherwise known as the y-intercept.

    m=(y2-y1)/(x2-x1)

    b=(y1*x2-y2*x1)/(x2-x1)

    In this case:

    x1=1/6
    x2=5/6
    y1= -1/3
    y2=5

    m=(y2-y1)/(x2-x1)
    m=[5-(-1/3)]/[(5/6)-(1/6)]
    m=[5+(1/3)]/(4/6)
    m=[(15/3)+(1/3)]/(4/6) Becouse 5=15/3
    m(16/3)/(4/6)
    m=(16*6)/(3*4)
    m=96/12

    m=8

    b=(y1*x2-y2*x1)/(x2-x1)
    b=[(-1/3)*(5/6)-((5*(1/6)]/[(5/6)-(1/6)]
    b=[(-5/18)-(5/6)]/(4/6)

    b=[(-5/18-(15/18)]/(4/6)

    Becouse 5/6=15/18

    b=(-20/18)/(4/6)
    b=(-20*6)/(4*18)
    b= -120/72
    b=(-12*10)(12*6)
    b=(-10/6)

    b= -5/3 Becouse (-10/6)= -5/3

    y=mx+b

    y=8x-(5/3)

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