Answer:
[tex]y=\frac{x}{0.3}-18 \ or \ 3y=10x-54[/tex]
is the equation of perpendicular line.
Step-by-step explanation:
We are given , the equation of a line is [tex]y=-0.3x. +6[/tex]
We can deduce that the slope of the given line is [tex]m=-0.3[/tex]
The slope of its perpendicular line would be , [tex]m_p= \frac{-1}{m} =\frac{1}{0.3}[/tex]
Now using the point slope form, [tex]y-y1= m_p(x-x1)[/tex]
Use the given point [tex](3,-8) \ as\ (x1,y1)[/tex] substituing these.
[tex]y-(-8)= \frac{1}{0.3} (x-3)\\y+8=\frac{x}{0.3} -10\\y=\frac{x}{0.3} -18[/tex]
Therefore the required equation of a line that is perpendicular to y=-0.3x. +6 and that passes through the point (3,-8) is y=\frac{x}{0.3} -18[/tex]
