Answer :
The line that passes through (-8,2) and is perpendicular to a line with a slope of -8 has the point-slope form x -8y +24 = 0
The point slope form of the line that passes through the points ([tex]x_{1} ,y_{1}[/tex]) and perpendicular to the line with a slope of “m” is given as,
[tex]y-y_{1} =-\frac{1}{m} (x-x_{1} )[/tex] (Equation-1)
Where “m” is the slope of the line. 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[/tex]
[tex]y_{1} =2[/tex]
We can substituting the values in equation-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}{2} (x+8)[/tex]
We can cross multiplying,
8y - 16 = (x+8)
8y – 16 = x +8
We can on rearranging ,
x -8y +16 + 8 = 0
x -8y +24 = 0
Hence,
The point slope form of given line is x -8y +24 = 0
Therefore,
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
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