Show that the parametric equation $ x=x_1+(x_2-x_1)t , y=y_1+(y_2-y_1)t$

Question

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

Answer 1

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.

Answer 2

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 $P_2$.