If from the continuous equation of the straight line we operate and group terms we obtain: $$$\displaystyle \begin{array}{rcl} \frac{x-p_1}{v_1} &=&\frac{y-p_2}{v_2} \\ v_2(x-p_1) &=& v_1(y-p_2)\\ v_2 \cdot x-v_2 \cdot p_1&=& v_1 \cdot y-v_1\cdot p_2 \\ v_2\cdot x-v_1 \cdot y+(v_1\cdot p_2-v_2\cdot p_1)&=&0 \\ Ax+By+C&=& 0\end{array}$$$
Where obviously,$$$\begin{array}{rcl}A&=&v_2 \\ B&=& -v1\\ C&=& v_1\cdot p_2 - v_2 \cdot p_1 \end{array}$$$An interesting property of this equation is that $$\overrightarrow {v}=(-B,A)$$ is a vector director of the straight line, and therefore $$\overrightarrow{w}=(A,B)$$ is a vector perpendicular to the straight line.
Find the implicit equation of the straight line $$r$$:$$$\displaystyle \frac{x-3}{-5}=\frac{y-4}{2}$$$
Computing and changing all the terms to one side we obtain: $$$\begin{array}{rcl} 2(x-3) &=& -5 (y-4) \\ 2x-6 &=& -5y+20 \\ 2x+5y-6-20 &=& 0 \\ 2x+5y-26&=&0\end{array}$$$
Therefore the implicit equation is $$2x + 5y - 26 = 0$$ and the vector $$\overrightarrow{v} = (-5, 2)$$ is a vector director of the straight line.