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: $\begin{array}{rcl} \frac{x - p_{1}}{v_{1}} & = & \frac{y - p_{2}}{v_{2}} \\ v_{2} (x - p_{1}) & = & v_{1} (y - p_{2}) \\ v_{2} \cdot x - v_{2} \cdot p_{1} & = & v_{1} \cdot y - v_{1} \cdot p_{2} \\ v_{2} \cdot x - v_{1} \cdot y + (v_{1} \cdot p_{2} - v_{2} \cdot p_{1}) & = & 0 \\ A x + B y + C & = & 0 \end{array}$

Where obviously, $\begin{array}{rcl} A & = & v_{2} \\ B & = & - v 1 \\ C & = & v_{1} \cdot p_{2} - v_{2} \cdot p_{1} \end{array}$ An interesting property of this equation is that $\vec{v} = (- B, A)$ is a vector director of the straight line, and therefore $\vec{w} = (A, B)$ is a vector perpendicular to the straight line.

Example

Find the implicit equation of the straight line $r$ : $\frac{x - 3}{- 5} = \frac{y - 4}{2}$

Computing and changing all the terms to one side we obtain: $\begin{array}{rcl} 2 (x - 3) & = & - 5 (y - 4) \\ 2 x - 6 & = & - 5 y + 20 \\ 2 x + 5 y - 6 - 20 & = & 0 \\ 2 x + 5 y - 26 & = & 0 \end{array}$

Therefore the implicit equation is $2 x + 5 y - 26 = 0$ and the vector $\vec{v} = (- 5, 2)$ is a vector director of the straight line.