Question

How do you write an equation of a line that contains the given point (-5,5) and is perpendicular to the given line y=-5x+9?

Answer 1

`(y - 5) = 1/5(x + 5)`

or

`y = 1/5x + 6`

Explanation:

To find the equation of the line perpendicular to the given line and going through the given point we will use the point-slope formula. We have been given a point so what we are missing is the slope.

The given line is in slope-intercept form.

The slope-intercept form of a linear equation is:

`y = color(red)(m)x + color(blue)(b)`

Where `color(red)(m)` is the slope and `color(blue)(b` is the y-intercept value.

Therefore we know the slope of the given line, `color(red)(m)` is `color(red)(-5)` - it is the coefficient of the `x` term.

A perpendicular line will have a slope which is the negative inverse of this line, or `color(red)(m = - 1/-5 = 1/5`.

We can now use the point-slope formula to write the equation for the perpendicular line.

The point-slope formula states: `(y - color(red)(y_1)) = color(blue)(m)(x - color(red)(x_1))`

Where `color(blue)(m)` is the slope and `color(red)(((x_1, y_1)))` is a point the line passes through.

Substituting the point we were given and the slope we calculated gives:

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

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

or, we can solve for `y` to convert to the more familiar slope-intercept form:

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

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

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

`y - 0 = color(blue)(1/5)x + 6`

`y = color(blue)(1/5)x + 6`