What will be the slope of $BC$? The vertex $A$ of triangle $\triangle ABC$ is $(3,-1)$. The equation of median $BE$ is $$6x+10y-59=0$$ and angle bisector $CF$ is $$x-4y+10=0.$$ Then what is the slope of $BC$?
Let slopes of $AC, CF, BC$ be $m1, m2, m3$ respectively, then
$$(m1-m2)/(1+m1*m2) = (m2-m3)/(1+m2*m3)$$
How to find $m1$?

Let's begin at point $C$.
As $C \in CF$ it follows that
$$C=(4y_C-10, y_C).\quad (1)$$
As $E$ is midpoint of $AC$ then the coordinates of E are the arithmetic mean of coordinates of $A$ and $C$, as it is shown in equation below:
$$E=(\frac{4y_C-7}{2}, \frac{y_C-1}{2}).\quad (2)$$
We know that $E \in BE$, therefore
$$6x_E+10y_E-59=0. \quad (3)$$
From $(2)$ and $(3)$ we get
$$y_C=5.$$
So we already know the location of three points:
$$A = (3, -1),$$
$$C = (10, 5),$$
$$E = (\frac{13}{2}, 2).$$
Let $K$ such that $\{K\}=CF \cap BE$, then
$$\left \{
\begin{array}{l}
x_K -4 y_K +10 =0\\
6x_K +10y_K -59 =0\\
\end{array}
\right. \Rightarrow K = (4, \frac{7}{2}).$$
We know that the slope of a stragiht line defined by PQ can be calculated by
$$m_{PQ}=\frac{y_P - y_Q}{x_P-x_Q}.$$
Let's then calculate $m_1$ and $m_2$ now:
$$m_1= m_{AC} = m_{CE} = \frac{3}{\frac{7}{2}} = \frac{6}{7} \quad (4)$$
$$m_2= m_{CF} = m_{CK} = \frac{\frac{3}{2}}{6} = \frac{1}{4} \quad (5)$$
Let $\gamma = m(\angle KCE)$ then
$$\tan \gamma = \frac{m_1-m_2}{1+m_1m_2}. \quad (6)$$
From $(4)$, $(5)$ and $(6)$ we get
$$\tan \gamma = \frac{1}{2}$$
As $m(\angle BCK)= m( \angle KCE)$ it follows that
$$ \frac{1}{2} = \frac{m_2-m_3}{1+m_2m_3} \Rightarrow \frac{1}{2} = \frac{\frac{1}{4}-m_3}{1+\frac{1}{4}m_3} \Rightarrow $$
$$\Rightarrow \frac{1}{2} + \frac{1}{8}m_3=\frac{1}{4} - m_3 \Rightarrow m_3= -\frac{2}{9}.$$
