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

`y=(x)/(0.3)-18 \ or \ 3y=10x-54`

is the equation of perpendicular line.

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= (-1)/(m) =(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)= (1)/(0.3) (x-3)\\y+8=(x)/(0.3) -10\\y=(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]