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
