Question

How do you find the slope that is perpendicular to the line `x = -5y - 5`?

Answer 1

See a solution process below:

Explanation:

First, let's put the equation in standard linear form. The standard form of a linear equation is: `color(red)(A)x + color(blue)(B)y = color(green)(C)`

Where, if at all possible, `color(red)(A)`, `color(blue)(B)`, and `color(green)(C)`are integers, and A is non-negative, and, A, B, and C have no common factors other than 1

The slope of an equation in standard form is: `m = -color(red)(A)/color(blue)(B)`

`x = -5y - 5`

`x + color(red)(5y) = -5y + color(red)(5y) - 5`

`x + 5y = 0 - 5`

`color(red)(1)x + color(blue)(5)y = color(green)(-5)`

Substituting for `color(red)(A)` and `color(blue)(B)` gives the slope of this line as:

`m = -color(red)(1)/color(blue)(5)`

Now, let's call the slope for the line perpendicular to this `m_p`

The rule of perpendicular slopes is:

`m_p = -1/m`

Substituting the slope we calculated gives:

`m_p = (-1)/(-1/5) = 5`

Answer 2

Slope of line is `5`

Explanation:

The slope of the line `x= -5y -5 or 5y = -x -5 or y = -x/5 -1` is

`m_1= -1/5` [Compare with standard slope intercept form

`y=mx+c`] . We know the product of slopes of perpendicular

lines is `m*m_1=-1 :. m= -1/m_1= -1/(-1/5)= 5`

Slope of line perpendicular to the line `x= -5y -5` is `5` [Ans]