Question

What is the equation of the line normal to `f(x)=1/abs(x-6)` at `x=1`?

Answer 1

Or in general form:

`25x+y-126/5 = 0`

Explanation:

Unpack the definition of `f(x)` as:

`f(x) = { (1/(x-6), if x > 6), (-1/(x-6), if x < 6) :}`

So:

`f'(x) = { (-1/(x-6)^2, if x > 6), (1/(x-6)^2, if x < 6) :}`

In particular:

`f(1) = 1/5`

`f'(1) = 1/25`

So the slope of the tangent at `(1, 1/5)` is `1/25` and the perpendicular normal must have slope `-1/(1/25) = -25`

The equation of the normal can be written in point slope form as:

`y - 1/5 = -25 (x - 1)`

Hence:

`y = -25x+25+1/5`

That is:

`y = -25x + 126/5`

in slope intercept form.

Or in general form:

`25x+y-126/5 = 0`