Write the point-slope form of the line that passes through (6,1) and is parallel to a line with a slope of -3 include all of your work in your final answer.

Answer: The point-slope form of the line that passes through (6,1) and is parallel to a line with a slope of -3 is 3x + y – 19 = 0 Solution: The point slope form of the line that passes through the points [tex]\left(x_{1} y_{1}\right)[/tex] and parallel to the line with slope “m” is given as  [tex]y - y_{1} = m\left(x - x_{1}\right)[/tex] --- eqn 1 Where “m” is the slope of the line. [tex]x_{1} \text { and } y_{1}[/tex]are the points that passes through the line. From question, given that slope “m” = -3 Given that the line passes through the points (6,1).Hence we get [tex]x_{1} = 6 ; y_{1} = 1[/tex]By substituting the values in eqn 1, we get the point slope form of the line which is parallel to the line having slope -3 can be found out. y – 1 = -3(x – 6) y – 1 = -3x +18 On rearranging the terms, we get 3x + y -1 – 18 = 0 3x + y – 19 = 0 Hence the point slope form of given line is 3x + y – 19 = 0