Write the point-slope form of the line that passes through (-8,2) and is perpendicular to a line with a slope of -8 include all of your work in your final answer.

Answer: The point-slope form of the line that passes through (-8,2) and is perpendicular to a line with a slope of -8 is x -8y +24 = 0 Solution: The point slope form of the line that passes through the points [tex]\left(x_{1} y_{1}\right)[/tex] and perpendicular to the line with a slope of “m” is given as  [tex]y-y_{1} = -\frac{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” = -8 Given that the line passes through the points (-8,2).Hence we get [tex]x_{1 } = -8 ; y_{1} = 2[/tex]By substituting the values in eqn 1 , we get the point slope form of the line which is perpendicular to the line having slope -8 can be found out. [tex]y - 2 = \frac{1}{8}(x + 8)[/tex]On cross multiplying we get, 8y - 16 = (x+8) 8y – 16 = x +8 On rearranging we get, x -8y +16 + 8 = 0 x -8y +24 = 0 hence the point slope form of given line is x -8y +24 = 0