Question

How do you write an equation of a line passing through (3, 2), perpendicular to `y=5x + 2`?

Answer 1

Find the negative reciprocal of the slope and, using our point, write the equation in slope-intercept form.

Explanation:

The negative reciprocal of the slope of a line is the slope of the line perpendicular to it. If our original slope is `5`, then our new slope is `-1/5`.

Now, let's plug in our value and solve!

`2=-1/5(3)+b`
`2=-3/5+b`
`2+3/5=b`
`13/5=b`

`y=-1/5x+13/5`

Answer 2

I got: `y=-1/5x+13/5`

Explanation:

Given our line:

`y=5x+2`

the slope `m` will be numerical coefficient of `x` that is: `m=5`
We know that the slope `m'` of the perpendiclar to our line must be:

`m'=-1/m=-1/5`

and the equation of the line with slope `m'` and passing through our point of coordinates (`x_0,y_0`) will be:

`y-y_0=m'(x-x_0)`

in our case:

`y-2=-1/5(x-3)`

`y=-1/5x+3/5+2`

`y=-1/5x+13/5`