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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 $m1$?

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up vote 2 down vote accepted

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 $$ \frac{1}{2} = \frac{m_2-m_3}{1+m_2m_3} \Rightarrow \frac{1}{2} = \frac{\frac{1}{4}-m_3}{1+\frac{1}{4}m_3} \Rightarrow $$ $$\Rightarrow \frac{1}{2} + \frac{1}{8}m_3=\frac{1}{4} - m_3 \Rightarrow m_3= -\frac{2}{9}.$$

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