Finding the parametric form of a standard equation

Question

I need to find the parametric form of $3x - 2y + 10 = 0$.

I found that the parametric form for this equation could be :

\begin{align}x &= t\\ y &= 5 + \frac{3}{2} t \end{align}

I did this by making $x = t$ and then solving for $y$

However, there are two other parametric forms that I don't know how to get to, could you please help me figure out how to get to them. They are:

\begin{align} (1)\qquad x &= 2t\quad \text{ and }\quad y = 5 + 3t \\ (2)\qquad x &= 3t\quad \text{ and } \quad y = 5 + \frac{9}{2}t \end{align}

Any help with be greatly appreciated

Answer 1

Treat $x$ as free variable and solve for $y$ we get $y=(-3x-10)/2$, now you can let $x$ be whatever you like and get the corresponding $y$.

Answer 2

the parametric equation for line in space $$\frac{x-x_0}{v_x}=\frac{y-y_0}{v_y}=\frac{z-z_0}{v_z}=t$$

in plane $$\frac{x-x_0}{v_x}=\frac{y-y_0}{v_y}=t$$

the line in question is $$y=\frac{3}{2}x+5$$

now we can create a vector parallel to the line in many ways so that $(v_y/v_x=3/2)$

$$v=i+\frac{3}{2}j$$

or $$v=2i+3j$$ or $$v=3i+\frac{9}{2}j$$

now we wiil find the parametric equation if the $v=i+\frac{3}{2}j$

select a point on line (0,5) $$\frac{x-x_0}{v_x}=\frac{y-y_0}{v_y}=t$$ $$\frac{x-0}{1}=t$$ hence $$x=t$$ $$\frac{y-5}{3/2}=t$$ hence $$y=3/2t+5$$

you can get the other parametric equation by select another form of vector