Explicit equation of the straight line

We can begin by the implicit equation that will be:

$3x+2y-6=0$ general, implicit or Cartesian equation.

Isolating $y$ we have:

$y=-\dfrac{3}{2}x+3$ Explicit equation

Now, since we have the slope, $m=-\dfrac{3}{2}$, a vector director of the straight line can be $\overrightarrow{v_1}=(1,-3/2)$.

Multiplying by $-2$, or from the general equation of the straight line, we have $\overrightarrow{v_2}=(-2, 3)$ which is another vector director of the straight line (and it is always more comfortable to work with entire numbers).

Now a point of the straight line could be $x=2$, and substituting $y=-\dfrac{3}{2}\cdot2+3=0$ and therefore $(2,0)$ is a point of the straight line.

This way the vectorial equation is:

$(x,y)=(2,0)+k\cdot(-2,3)$ Vectorial equation

and now we can easily obtain the parametrical equations and the continuous equation:

$\begin{array}{c} x=2-2k \\ y=3k \end{array}$ Parametrical equations

and isolating $k$ and equaling them we have:

$\dfrac{x-2}{-2}=\dfrac{y}{3}$ Continuous equation

Finally , as have we already found a point of the straight line and the slope, the equation slope-point, for the above mentioned point coincides with the explicit equation. Another possibility would be to take the point $x=0$,

$y=-\dfrac{3}{2}\cdot0+3=3$ and then the slope-point equation of the straight line would be:

$y-3=-\dfrac{3}{2}x$ Equation slope-point