+1557 The line $y = mx$ bisects the angle between the two lines shown below. Find $m$.
+399 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 \(\bf{m=\frac{\sqrt{962}-1}{31}}\).\(\)
