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I am not sure how to do this one at all, I can't even start it.

I am suppose to show that $$x = x_1+(x_2-x_1)t \\ y= y_1+(y_2-y_1)t$$where t is between zero and one describes the line segment that joins $P_1(x_1, y_1)$ and $P_2(x_2,y_2)$

I really have no clue what to do and how to work with this many different variables.

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Do you want $x = x_1 + (x_2-x_1)t$? –  Nicholas Kirchner Jun 25 '12 at 0:43
    
@NicholasKirchner Yes that is what I meant, too late now I guess. –  user138246 Jun 25 '12 at 1:44
    
Jordan, the only variable there is $x$ (and $y(x)$, well). The other are fixed arbitrary constants. Don't get intimidated by it! –  Pedro Tamaroff Jun 25 '12 at 1:44

2 Answers 2

up vote 2 down vote accepted

There are three things to prove here:

  1. the parametrization begins at $(x_1,y_1)$,
  2. the parametrization ends at $(x_2,y_2)$,
  3. the parametrization is a line segment.

The first two are easy to prove: we have $0 \leq t \leq 1$, so plug $t=0$ into the equations for $x$ and $y$. You should obtain $x = x_1$ and $y = y_1$, which gives us item #1 on our todo list. Do the same for $t=1$, and you'll get item #2.

The third requires a bit of algebra, and it must be kept in mind that $x_1,x_2,y_1,y_2$ are all constants, while $x,y$ are variables. Let us solve the $x$ equation for $t$: $$ x = x_1 + (x_2 - x_1)t $$ $$ \frac{x - x_1}{x_2-x_1} = t $$ Now, plug that into the $y$ equation: $$ y = y_1 + (y_2-y_1)t $$ $$ y = y_1 + (y_2-y_1)\frac{x - x_1}{x_2-x_1} $$

The goal here is to get it into the form $y = mx+b$, so the algebra will be focused on figuring out what's multiplying $x$ and what isn't. We'll start by breaking the fraction into two:

$$ y = y_1 + (y_2-y_1)\frac{x}{x_2-x_1} + (y_2-y_1)\frac{- x_1}{x_2-x_1} $$ Now, move the $y_1$ at the beginning to join it with the other stuff that does not depend on $x$: $$ y = (y_2-y_1)\frac{x}{x_2-x_1} + y_1 - (y_2-y_1)\frac{x_1}{x_2-x_1} $$ From here, we can see that we do have a line if we set $$ m =\frac{y_2-y_1}{x_2-x_1} $$ $$ b = y_1 - (y_2-y_1)\frac{x_1}{x_2-x_1} $$

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You could have stopped at $$y = y_1 + (y_2-y_1)\frac{x - x_1}{x_2-x_1}$$ That is a line with slope $\dfrac{y_2-y_1}{x_2-x_1}$ which passes through $(x_1, y_1)$ –  Pedro Tamaroff Jun 25 '12 at 1:25
    
@Peter: Yeah, but the original questioner indicated discomfort with so many variables. –  Nicholas Kirchner Jun 25 '12 at 1:40

The equation of line between two point is given by $$ y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$$ $$ \implies \frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1} = t \text{ (let) } $$ $$ \implies y = y_1 + t(y_2 - y_1) \text{ and } \implies x = x_1 + t(x_2 - x_1) $$

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