How to find the coordinates where the altitude of a triangle intersects the base? I have a triangle with known, but random coordinates for each point.
Let's assume A(3,4), B(5,7), C(13.5,8.5)

How can I find the coordinates where the altitude from point B intersects the AC segment?
I've calculated the length of BD, AD and DC segments, the slope of AC but I'm not sure what's next.
 A: The line $AC$ can be denoted by the equation $y = \frac{3}{7}x + \frac{19}{7}$. 
The altitude is perpendicular to $AC$, so it will have slope $-\frac{7}{3}$ and go through $(5,7)$, so it is the line $y' = -\frac{7}{3}x + \frac{56}{3}$.
Now just solve those equations:
$$\begin{split}
y &= \frac{3}{7}x + \frac{19}{7} \\
y &= -\frac{7}{3}x + \frac{56}{3}
\end{split}$$
A: The line $AC$ is defined by $$\frac{y-y_A}{x-x_A}=\frac{y_C-y_A}{x_C-x_A}$$   
Then we know $AC(x)=m_{AC}x+b_{AC}$, where $m_{AC}$ is the slope and $b_{AC}$ is the y-intercept (both are known).
Let $BD(x)=m_{BD}x+b_{BD}$. Then $m_{BD}=\frac{-1}{m_{AC}}$ and $b_{BD}=y_B-m_{BD}x_B$.   
Now solve $AC(x)=BD(x)$.
A: $$
x_? = \frac{x_A \dfrac{y_C - y_A}{x_C - x_A} + y_B + x_B \dfrac{x_C - x_A}{y_C - y_A}}{\dfrac{y_C - y_A}{x_C - x_A} + \dfrac{x_C - x_A}{y_C - y_A}},\\
y_? = \frac{y_C - y_A}{x_C - x_A} \left( \frac{x_A \dfrac{y_C - y_A}{x_C - x_A} + y_B + x_B \dfrac{x_C - x_A}{y_C - y_A}}{\dfrac{y_C - y_A}{x_C - x_A} + \dfrac{x_C - x_A}{y_C - y_A}} - x_A \right).
$$
