Equation of the line which is passing through the points [tex](-1,7)[/tex] and perpendicular to the line [tex]$y=-\frac{1}{8}x-1$[/tex]
is [tex]y=8x+15[/tex]
How to find the equation which passing through one point [tex](-1,7)[/tex]and perpendicular to the equation [tex]$y=-\frac{1}{8}x-1$[/tex] ?
We know that the equation of line in form of slope and point is
[tex]y-y_{1} =m(x-x_{1} )[/tex]
So first equation of line which is passing through point [tex](-1,7)[/tex] is
[tex]y-7=m_{1} (x-(-1))\\y-7=m_{1}(x+1)[/tex].....(a)
Line [tex]$y=-\frac{1}{8}x-1$[/tex] perpendicular to the line (a)
And we know when two lines are perpendicular then
[tex]m_{1} m_{2} =-1[/tex]
So [tex]m_{1} =-1/m_{2}[/tex]
and we know [tex]m_{2} =-1/8[/tex]
So
[tex]m_{1} =-1/\frac{-1}{8} \\=-1*8/-1\\=8[/tex]
The equation of the line is
[tex]y-7=8(x+1)\\y-7=8x+8\\y=8x+8+7\\y=8x+15[/tex]
Learn more about the equation of the line is here :
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