# Normal Line to a Tangent

## Key Questions

• If a tangent line has the equation

$y - {y}_{1} = m \left(x - {x}_{1}\right)$,

then the normal line at the point of contact is

$y - {y}_{1} = - \frac{1}{m} \left(x - {x}_{1}\right)$.

I hope that this was helpful.

• The normal line is the line that is perpendicular to the the tangent line.

If the slope of a line is $m$ then the slope of the perpendicular line is $- \frac{1}{m}$, this is also known as the negative reciprocal.

The given equation is $y = \frac{5}{6} x - 9$ the slope is $\frac{5}{6}$ so the slope of the normal is $- \frac{6}{5}$.

The point $\left(x , y\right) \to \left(4 , 8\right)$

$y = m x + b \to$ Substitute in the values of $m$, $x$ and $y$

$8 = - \frac{6}{5} \left(4\right) + b$

$8 = - \frac{24}{5} + b$

$\frac{24}{5} + 8 = b$

$\frac{24}{5} + \frac{40}{5} = b$

$\frac{64}{5} = b$

The equation of the normal line is $\to y = - \frac{6}{5} x + \frac{64}{5}$

• A normal line is the line perpendicular to a tangent line at the point of contact.