Coordinates

We have a very interesting property, and very useful property when it comes to slope: $\tan=\frac{O}{A}$, opposite over adjacent, but when put in the coordinate plane, that is $\tan=\frac{rise}{run}$, and what else is rise over run? Slope! .

Set the angle formed by y=5x and the x axis as $\theta$, and the angle formed by y=1/6x and the x axis as $\phi$.

We want to find $\tan(\phi+\frac{\theta-\phi}{2})=\tan(\frac{\theta+\phi}{2})$.

Use some of our identities:

With a lot of use of tangent identities, we know $\tan(\theta)=5, \tan(\phi)=1/6$.

Directly applying our tangent angle identity: $\tan(\theta+\phi)=\frac{5+\frac{1}{6}}{1-\frac{5}{6}}=31$.

Calculate, $\tan(\theta+\phi)=\frac{\sin(\theta+\phi)}{1+\cos(\theta+\phi)}$.

Draw a triangle and calculate sin and cos.

$\sin(\theta+\phi)=\frac{31}{\sqrt{962}}$, $\cos(\theta+\phi)=\frac{1}{\sqrt{962}}$

Substituting in, we get $y=(\frac{\sqrt{962}-1}{31})x$ as our line, so $\bf{m=\frac{\sqrt{962}-1}{31}}$.