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    x1+y*v1 = x2+y*v2

    =>(x1-x2)+y(v1-v2) = 0

    =>(x1-x2) = -y(v1-v2) Removing the '-ve' sign from RHS =>(x2-x1) = y(v1-v2)

    =>(x2-x1)/(v1-v2) = y ----(1)

    If you multiply -1 to both the numerator and denominator, then

    =>(x1-x2)/(v2-v1) = y ----(2)

    Thus equation (1) and (2) are the same.