Problems from Explicit equation of the straight line

Development:

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

Solution:

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

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

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

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

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

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

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