Question

Write the equation of a line that is perpendicular to y=-0.3x. +6 and that passes through the point (3,-8)

Answer 1

Answer:

$y=\frac{x}{0.3}-18 \ or \ 3y=10x-54$

is the equation of perpendicular line.

Step-by-step explanation:

We are given , the equation of a line is  $y=-0.3x. +6$

We can deduce that the slope of the given line is $m=-0.3$

The slope of its perpendicular line would be , $m_p= \frac{-1}{m} =\frac{1}{0.3}$

Now using the point slope form, $y-y1= m_p(x-x1)$

Use the given point $(3,-8) \ as\ (x1,y1)$ substituing these.

$y-(-8)= \frac{1}{0.3} (x-3)\\y+8=\frac{x}{0.3} -10\\y=\frac{x}{0.3} -18$

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]