# What is the slope of the line perpendicular to  y=16/7x-11?

Jul 18, 2016

$- \frac{7}{16}$

#### Explanation:

Recall that the slope $m$ of a straight line is related to the angle $\theta$ that it makes with the $X$ axis by

$m = \tan \theta$

Two mutually perpendicular lines obey ${\theta}_{2} = {\theta}_{1} + \frac{\pi}{2}$ so that

${m}_{1} {m}_{2} = \tan {\theta}_{1} \tan {\theta}_{2} = \tan {\theta}_{1} \tan \left({\theta}_{1} + \frac{\pi}{2}\right) = \tan {\theta}_{1} \left(- \cot {\theta}_{1}\right) = - 1$

Jul 18, 2016

$- \frac{7}{16}$

#### Explanation:

Consider the standard equation form $y = m x + c$ where $m$ is the gradient.

Any line perpendicular to this has the gradient of $- \frac{1}{m}$

So for $y = \frac{16}{7} x - 11$ the gradient of a line perpendicular to this is

$- \frac{7}{16}$