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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

Corvo
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    What are you asking for a parametrization of? The line segments inbetween each vertex? The region enclosed by the triangle? The vertices themselves? If asking about the line segments, are you curious about each individually, or are you going to travel along one after another in a continuous path? If so, what orientation: clockwise or counterclockwise? – JMoravitz Mar 16 '15 at 17:23
  • Like segment it is – Corvo Mar 16 '15 at 18:49
  • Line segment. Wrote it wrong – Corvo Mar 16 '15 at 18:50
  • The line segment between any two points: $(x_1,y_1)$ and $(x_2,y_2)$ can be described as $(x_1 + (x_2-x_1)t, y_1 + (y_2-y_1)t):~0\leq t\leq 1$. – JMoravitz Mar 16 '15 at 18:52
  • It says a(1-t)b = x, bt - a = y – Corvo Mar 16 '15 at 19:59
  • I don't understand how one would get such answer – Corvo Mar 16 '15 at 20:00

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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:

$$[x,y] = \begin{cases} [0+3at,0] & 0\leq t\leq \frac{1}{3}\\ [2a-3at,3bt-b] & \frac{1}{3}<t\leq\frac{2}{3}\\ [0,3b-3bt] & \frac{2}{3}<t\leq 1\end{cases}$$

JMoravitz
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