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

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Answer 1 · 2016-01-02T13:12:20

Or in general form:

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

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#