Parametrisize of vertecies (0,0) (a,0) (0,b) Hi i have been given 3 vertecies. (0,0) (a,0) (0,b)
The constants a and b are >=0. This forms a backwards triangle. The parametisation don't make sense to me, so basically what i am asking is for someone to explain it to me in a way that makes sense to me.
Sorry for bad english 
 A: Something first to notice, is that if an equation is of the form:
$$p_0 + mt = p$$
where $p_0$ and $m$ are regular (unchanging) numbers and not variables, then it is referred to as being "linear" and will be represented by a line.  By putting conditions on $t$ such as $t_i\leq t\leq t_f$, where $t_i$ and $t_f$ are the initial and final times and are unchanging numbers, then the equation will represent a line segment.  (this should remind you of the equation of a line in point-slope form: $y=mx+b$)
Notice that when $m=(p_1-p_0)$ and $0\leq t\leq 1$, at $t=0$ you have:
$$p_0 + (p_1-p_0)t = p_0+(p_1-p_0)\cdot 0 = p_0$$
and at $t=1$ you have:
$$p_0+(p_1-p_0)t = p_0+(p_1-p_0)\cdot 1 = p_0 + p_1 - p_0 = p_1$$
Since $m=(p_1-p_0)$ is non-variable (unchanging) once $p_1$ and $p_0$ are specified, the equation $p_0 + (p_1-p_0)t$ with $0\leq t\leq 1$ is linear, and describes the line segment starting from $p_0$ and ending at $p_1$.
This generalizes to higher dimensions as well.  What I referred to as $p_0,p_1$ and $p$ could be $n$-tuples referring to points in $n$-dimensional space.
In your example, we are working in two dimensional space.  Looking at the hypotenuse of the triangle in particular and letting $p_0 = [a,0]$ and $p_1 = [0,b]$, we get:
$$[x,y] = [a,0] + ([0,b]-[a,0])t~~~~~~0\leq t\leq 1\\ = [a-at,bt]$$
Here, I am using vector notation, where I am describing both what happens to $x$ and what happens to $y$ in the same line.  By equating the entries of the vector on the left with the corresponding entry on the right, you could have written this as:
$$\begin{cases} x = a-at & \\
 & 0\leq t \leq 1\\
y = bt & \end{cases}$$
This is a parametric equation describing the line segment.  Please notice also that just as there are infinitely many ways to describe a line, there are infinitely many ways to parametrize a curve.
Even more generally, if you don't wish to use $t_i=0$ and $t_f=1$, you may parametrize a line segment from arbitrary starting time $t_i$ to arbitrary final time $t_f$ as:
$$p = p_0 + \frac{p_1-p_0}{t_f-t_i}(t-t_i)~~~~~~~t_i\leq t\leq t_f$$
which again comes from the point-slope form for lines.  (notice that when $t_i=0$ and $t_f=1$ the equation here agrees with the one above and by plugging in $t=t_i$ and $t=t_f$ you get the initial and final points respectively).
As such, we are able to parametrize the entire path around the triangle: (done here as counter-clockwise starting from the origin with $0\leq t\leq 1$ and each line segment taking precisely one third the time)
$$[x,y] = \begin{cases} [0,0] + \frac{[a,0]-[0,0]}{\frac{1}{3}-0}(t-0) & 0\leq t\leq \frac{1}{3}\\
[a,0] + \frac{[0,b] - [a,0]}{\frac{2}{3}-\frac{1}{3}}(t-\frac{1}{3}) & \frac{1}{3}<t\leq\frac{2}{3}\\
[0,b] + \frac{[0,0]-[0,b]}{1-\frac{2}{3}}(t-\frac{2}{3}) & \frac{2}{3}<t\leq 1\end{cases}$$
which simplifies to:
$$[x,y] = \begin{cases} [0+3at,0] & 0\leq t\leq \frac{1}{3}\\
[2a-3at,3bt-b] & \frac{1}{3}<t\leq\frac{2}{3}\\
[0,3b-3bt] & \frac{2}{3}<t\leq 1\end{cases}$$
