Chapter 1.4, Problem 48E

(a)

To determine

To show: The parameterization for the line that passes through the points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is $x=x_1+\left(x_2-x_1\right)t$ , $y=y_1+\left(y_2-y_1\right)t$ , in the interval $-\infty <t<\infty$.

Explanation of Solution

Given information: The points are $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$.

Proof:

Solve the equation $x=x_1+\left(x_2-x_1\right)t$ for $t$.

  $\begin{array}{c}x=x_1+\left(x_2-x_1\right)t\\ \left(x_2-x_1\right)t=x-x_1\\ t=\frac{x-x_1}{x_2-x_1}\end{array}$

Solve the equation $y=y_1+\left(y_2-y_1\right)t$ for $t$.

  $\begin{array}{c}y=y_1+\left(y_2-y_1\right)t\\ \left(y_2-y_1\right)t=y-y_1\\ t=\frac{y-y_1}{y_2-y_1}\end{array}$

Therefore, $\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}$.

Multiply both sides of the equation $\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}$ by $y_2-y_1$.

  $\begin{array}{c}\frac{x-x_1}{x_2-x_1}\times \left(y_2-y_1\right)=\frac{y-y_1}{y_2-y_1}\times \left(y_2-y_1\right)\\ \frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)=y-y_1\\ y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)\end{array}$

Therefore, it is proved that $x=x_1+\left(x_2-x_1\right)t$ , $y=y_1+\left(y_2-y_1\right)t$ , $-\infty <t<\infty$ is parameterization for the line through the points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$.

(b)

To determine

To find: The parameterization for the line segment with endpoints $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$.

Answer to Problem 48E

The parameterization for the line segment with endpoints $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is $x=x_1+\left(x_2-x_1\right)t$ , and $y=y_1+\left(y_2-y_1\right)t$ for $0\le t\le 1$.

Explanation of Solution

Given information: The points are $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$.

Calculation:

In the problem one possible parameterization for a line through points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is $x=x_1+\left(x_2-x_1\right)t$ , and $y=y_1+\left(y_2-y_1\right)t$ . So, it is only required to find a correct parameter interval to satisfy the endpoints.

Since the initial point is $\left(x_1,y_1\right)$.

Substitute 0 for $t$ in the equations $x=x_1+\left(x_2-x_1\right)t$ , and

  $y=y_1+\left(y_2-y_1\right)t$.

  $\begin{array}{c}x=x_1+\left(x_2-x_1\right)\left(0\right)\\ =x_1\\ y=y_1+\left(y_2-y_1\right)\left(0\right)\\ =y_1\end{array}$

The terminal point is $\left(x_2,y_2\right)$.

Substitute 1 for $t$ in the equations $x=x_1+\left(x_2-x_1\right)t$ , and $y=y_1+\left(y_2-y_1\right)t$.

  $\begin{array}{c}x=x_1+\left(x_2-x_1\right)\left(1\right)\\ =x_2\\ y=y_1+\left(y_2-y_1\right)\left(1\right)\\ =y_2\end{array}$

So, the parametric interval is $0\le t\le 1$.

Therefore, the parameterization for the line segment with endpoints $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is $x=x_1+\left(x_2-x_1\right)t$ , and $y=y_1+\left(y_2-y_1\right)t$ for $0\le t\le 1$.