TITLE: The slope of a triangle's Euler line in terms of the slopes of its sides
QUESTION [1 upvotes]: Let $p$, $q$, $r$, be the slopes of sides of a triangle, and $m$ the slope of Euler line of the triangle. Then we may say 
$$m= -\frac{3+pq+pr+qr}{p+q+r+3pqr}$$
provided that the denominator isn't zero. If it's zero the euler line of the triangle is parallel to $y$ axis.
I would like hints for proving this formula.

REPLY [0 votes]: Let the slopes of the triangle's sides be $m_i:=\tan\theta_i$ (with $i=1,2,3$), and define $\sigma :=\theta_1+\theta_2+\theta_3$.
The Euler line passes through a triangle's circumcenter and centroid. If we inscribe the triangle in the origin-centered unit circle, the $y$-over-$x$ ratio of the centroid's coordinates give the line's slope. Specifically, a triangle with vertices 
$$P_i := (\;\sin(2\theta_i-\sigma), \cos(2\theta_i-\sigma)\;)$$ 
has centroid $\frac13(P_1+P_2+P_3)$, so that the slope of its Euler line is
$$\begin{align}
&\phantom{=\;-} \frac{\cos(\theta_1-\theta_2-\theta_3)+\cos(-\theta_1+\theta_2-\theta_3)+\cos(-\theta_1-\theta_2+\theta_3)}{\sin(\theta_1-\theta_2-\theta_3)+\sin(-\theta_1+\theta_2-\theta_3)+\sin(-\theta_1-\theta_2+\theta_3)}\\[6pt]
&= -\frac{3\ddot{\theta_1}\ddot{\theta_2}\ddot{\theta_3}+\ddot{\theta_1}\dot{\theta_2}\dot\theta_3+\dot\theta_1\ddot\theta_2\dot\theta_3+\dot\theta_1\dot\theta_2\ddot\theta_3}{3\dot\theta_1\dot\theta_2\dot\theta_3+\dot\theta_1\ddot\theta_2\ddot\theta_3+\ddot\theta_1\dot\theta_2\ddot\theta_3+\ddot\theta_1\ddot\theta_2\dot\theta_3} \quad\left(\;\dot\theta:=\sin\theta, \;\ddot\theta:=\cos\theta\;\right)\\[6pt]
&=-\frac{\ddot\theta_1\ddot\theta_2\ddot\theta_3\left(3+m_2m_3+m_3m_1+m_1m_2\right)}{\ddot\theta_1\ddot\theta_2\ddot\theta_3\left(3m_1m_2m_3+m_1+m_2+m_3\right)} \\[6pt]
&=-\frac{3+m_2m_3+m_3m_1+m_1m_2}{3m_1m_2m_3+m_1+m_2+m_3} 
\end{align}$$
Now, we need only note that the slope of $\overline{P_2P_3}$ is
$$\frac{\cos(-\theta_1+\theta_2-\theta_3)-\cos(-\theta_1-\theta_2+\theta_3)}{\sin(-\theta_1+\theta_2-\theta_3)-\sin(-\theta_1-\theta_2+\theta_3)} =\frac{2\sin(\theta_2-\theta_3)\sin\theta_1}{2\sin(\theta_2-\theta_3)\cos\theta_1} = m_1$$
Likewise, $\overline{P_3P_1}$ and $\overline{P_1P_2}$ have respective slopes $m_2$ and $m_3$. $\square$
{"set_name": "stack_exchange", "score": 1, "question_id": 2912551}