Parametric equations and line segments 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.
 A: There are three things to prove here:


*

*the parametrization begins at $(x_1,y_1)$,

*the parametrization ends at $(x_2,y_2)$,

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