Continuous equation of the straight line

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 $r$: $$\displaystyle \frac{x-3}{-5}=\frac{y-4}{2}$$