# What is the equation of the line normal

What is the equation of the line normal to $f\left(x\right)=\frac{1}{x-2}$ at x=1?
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Quimpoah3b
Explanation:
first find f'(x)
$f\left(x\right)={\left(x-2\right)}^{-1}$
${f}^{\prime }\left(x\right)=-{\left(x-2\right)}^{-2}$
then find the slope of the tangent line at x=1, which is also f'(1)
${f}^{\prime }\left(1\right)=-{\left(1-2\right)}^{-2}$
${f}^{\prime }\left(1\right)=-{\left(-1\right)}^{-2}$
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 $f\left(1\right)=\frac{1}{1-2}=\frac{1}{-1}=-1$
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

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