Question

The equation of line AB is (y-3)= 5 (x-4). What is the slope of a line perpendicular to line AB?

Answer 1

`-1/5`

Explanation:

By definition, two perpendicular lines, say `barX` and `barY`, have slopes such that `m_Y=1/-m_X`.

To simplify, just take the negative reciprocal of one line to get the slope of a perpendicular line.

In your case, the slope of `bar(AB)` is `5`, as detailed by the given point-slope equation. If you're confused as to why this is, refer to the bottom of this answer.

If we plug `5` into the equation above, we can see:

`m_(AB')=1/-m_(AB)`
`m_(AB')=1/-(5)`
`m_(AB')=-1/5`

Assume `bar(AB')` is the line perpendicular to `bar(AB)`.

Here I'll explain how I got `5` as the slope of the line from this equation:

The general form of point-slope form is:

`(y-y_1)=m(x-x_1)`

Where
`y_1` is the `y`-value of a point on the line,
`x_1` is the `x`-value of the same point, and
`m` is the slope of the line.

Hence, I was able to figure out that in your case, the slope of your initial line was `5`. (I figured this might be helpful for people who didn't recognize point-slope form.)

Hope this helps!