What is the equation of the line normal to # f(x)=1/(x-2) # at # x=1#?

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Answer 1 · 2018-04-04T03:44:44

#y=x-2#

first find #f'(x)#
#f(x)=(x-2)^(-1)#
#f'(x)=-(x-2)^(-2)#

then find the slope of the tangent line at #x=1#, which is also #f'(1)#
#f'(1)=-(1-2)^(-2)#
#f'(1)=-(-1)^(-2)#
#f'(1)=-1#

use this fact: normal slope #*# tangent slope #= -1#

normal slope #*(-1)=-1#
normal slope #=1#

now find the point where #x=1#
the y-value is #f(1)=1/(1-2)=1/(-1)=-1#

for the normal line, the point is #(1,-1)# and the slope is 1

use point-slope form:
#y-(-1)=1(x-1)#
#y+1=x-1#
#y=x-2#