Continuous equation of the straight line

This is the result of isolating $k$ in the parametrical equations and making them equal: $\begin{array}{rcl} x & = & p_{1} + k \cdot v_{1} \\ y & = & p_{2} + k \cdot v_{2} \end{array}}$ $\frac{x - p_{1}}{v_{1}} k = \frac{y - p_{2}}{v_{2}} \frac{x - p_{1}}{v_{1}} = \frac{y - p_{2}}{v_{2}}$

Example

Find the continuous equation of the straight line $r$ that crosses the points $(3, 4)$ and $(- 2, 6)$ .

The vector equation with $A = (3, 4)$ and $B = (- 2, 6)$ is: $(x, y) = A + k \cdot \vec{A B} = (3, 4) + k \cdot (- 5, 2)$ Therefore, the parametrical equations of the straight line are: $\begin{array}{rcl} x = 3 - 5 \cdot k \\ y = 4 + 2 \cdot k \end{array}}$

We isolate $k$ : $\begin{array}{rcl} k & = & \frac{x - 3}{- 5} \\ k & = & \frac{y - 4}{2} \end{array}$ and we make them equal obtaining this way the continuous equation of the straight line $r$ : $\frac{x - 3}{- 5} = \frac{y - 4}{2}$