Show that the parametric equation $ x=x_1+(x_2-x_1)t , y=y_1+(y_2-y_1)t$ Can anyone help me to solve this?
Show that the parametric equation
$ x=x_1+(x_2-x_1)t $
$ y=y_1+(y_2-y_1)t\ $
with $(0\le t\le 1)$
describe the segment that joint the point $P_1=(x_1,y_1)$ and 
$P_2=(x_2,y_2)$
Thanks all
 A: Let us use the idea of vector. Any point $(x,y)$ on the segment can be represented as
$$x=x_1+(x_2-x_1)t, y=y_1+(y_2-y_1)t$$
where $0\le t\le 1$
because $(x_1,y_1)$ is the start point and $(x_2-x_1, y_2-y_1)$ is the orientation vector from $(x_1,y_1)$ to $(x_2,y_2)$.
Let $O(0,0), A(x_1,y_1),B(x_2,y_2), P(x,y)$ where $P$ represents any point on the segment.
$$\vec{OP}=\vec{OA}+t\vec{AB}=\vec{OA}+t(\vec{OB}-\vec{OA})$$
where $0\le t\le 1$. You'll see that this is the same representation as above.
A: What I am saying is not different from mathlove! In fact I like his answer, but I thought I will give you another way to look at it. See also Jonas12's comment.
If you put $t=0$ and $t=1$ you get the two endpoints.
If $P_1 = P_2$ then the parametric curve is just a point. 
Using calculus:
If $P_1 \ne P_2$ then both $\frac{dx}{dt}$ and $\frac{dy}{dt}$ are both constant and one of them is not zero.  $\frac{dy}{dx}$ is a constant if tou can divide by $\frac{dx}{dt}$ since 
$$
\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$
If $\frac{dx}{dt} = 0$ then $\frac{dx}{dy} = 0$, i.e. the curve is a vertical line.
Using algebra:
If $P_1 \ne P_2$, suppose that $x_1 \neq x_2$. Then we can solve for $t$ from the first equation:
$$
t=\frac{x-x1}{x_2-x_1}$$
Substitute in the second to get
$$
y = y_1 + \frac{y_2-y_1}{x_2-x_1} \left(x-x_1 \right) $$
This is the equation of the line connecting $P_1$ and $P_2$.
