Question 1150510
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The slope of the first line is


    m1 = {{{((-3)-2k)/(4-7)}}} = {{{(2k+3)/3}}}.


The slope of the second line is


    m2 = {{{(5-(k+1))/(3-1)}}} = {{{(4-k)/2}}}.


The line are parallel if and only if m1 = m2


    {{{(2k+3)/3}}} = {{{(4-k)/2}}}.


To solve this equation, first cross-multiply; then simplify


    2*(2k+3) = 3*(4-k)

    4k + 6 = 12 - 3k

    4k + 3k = 12 - 6

    7k      = 6

     k      = {{{6/7}}}.      <U>ANSWER</U>
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