General equation (or Cartesian or implicit) of the straight line

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.