This is the result of isolating $$k$$ in the parametrical equations and making them equal: $$$\left .\begin{array} {rcl} x & = & p_1+k\cdot v_1 \\ y & = & p_2+k \cdot v_2 \end{array}\right \}$$$ $$$\displaystyle \frac{x-p_1}{v_1} \\ k=\frac{y-p_2}{v_2} \\ \frac{x-p_1}{v_1}=\frac{y-p_2}{v_2}$$$
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: $$$(x, y) = A + k \cdot \overrightarrow {AB} = (3, 4) + k \cdot (-5, 2)$$$ Therefore, the parametrical equations of the straight line are: $$$\left. \begin{array}{rcl} x=3-5 \cdot k \\ y=4+2 \cdot k \end{array} \right\}$$$
We isolate $$k$$: $$$\displaystyle \begin{array}{rcl} k&=&\frac{x-3}{-5} \\ k &=& \frac{y-4}{2}\end{array}$$$ and we make them equal obtaining this way the continuous equation of the straight line $$r$$: $$$\displaystyle \frac{x-3}{-5}=\frac{y-4}{2}$$$