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

Mar 3, 2017

Equation of normal is $x - 2 y - 3 = 0$

#### Explanation:

The line normal to $f \left(x\right)$ at $x = {x}_{0}$ is perpendicular to the tangent at the point $\left({x}_{0} , f \left({x}_{0}\right)\right)$.

As slope of the tangent at $x = {x}_{0}$ is $f ' \left({x}_{0}\right)$, the slope of normal is $- \frac{1}{f ' \left({x}_{0}\right)}$ and equation of normal is $y - f \left({x}_{0}\right) = - \frac{1}{f ' \left({x}_{0}\right)} \left(x - {x}_{0}\right)$.

Here we are seeking normal at $x = 1$ and as $f \left(x\right) = \frac{x}{x - 2}$, at this point $f \left(1\right) = \frac{1}{1 - 2} = - 1$.

For slope let us work out derivative of $f \left(x\right) = \frac{x}{x - 2} = 1 + \frac{2}{x - 2}$

and $f ' \left(x\right) = - \frac{2}{x - 2} ^ 2$ and slope of tangent at $x = 1$ is $f ' \left(1\right) = - \frac{2}{1 - 2} ^ 2 = - 2$

The slope of normal is then $- \frac{1}{-} 2 = \frac{1}{2}$

and equation of tangent is $y - \left(- 1\right) = - 2 \left(x - 1\right)$ or $2 x + y - 1 = 0$ and that of

normal is $y - \left(- 1\right) = \frac{1}{2} \left(x - 1\right)$ or $2 y + 2 = x - 1$ or $x - 2 y - 3 = 0$
graph{(x-2y-3)(xy-x-2y)(2x+y-1)=0 [-10, 10, -5, 5]}