Chapter 0.4, Problem 48E

(a.)

To determine

To Show: If $a\ne 0$ , then the given expressions are a parametrization for the line through $\left(x_0,y_0\right)$ of slope $\frac{b}{a}$ .

Answer to Problem 48E

It has been shown that, if $a\ne 0$ , then the given expressions are a parametrization for the line through $\left(x_0,y_0\right)$ of slope $\frac{b}{a}$ .

Explanation of Solution

Given:

The expressions; $x=x_0+at$ , $y=y_0+bt$ , where $-\infty <t<\infty$ .

Concept used:

The equation of the line through $\left(x_0,y_0\right)$ of slope $\frac{b}{a}$ is given as $\frac{y-y_0}{x-x_0}=\frac{b}{a}$ .

Calculation:

It is given that $x=x_0+at$ , $y=y_0+bt$ , where $-\infty <t<\infty$ .

Simplifying,

  $x-x_0=at$ ,

  $y-y_0=bt$

Dividing the second equation by the first equation,

  $\frac{y-y_0}{x-x_0}=\frac{bt}{at}$

Simplifying,

  $\frac{y-y_0}{x-x_0}=\frac{b}{a}$

As discussed, this is the equation of the line through $\left(x_0,y_0\right)$ of slope $\frac{b}{a}$ .

Conclusion:

It has been shown that, if $a\ne 0$ , then the given expressions are a parametrization for the line through $\left(x_0,y_0\right)$ of slope $\frac{b}{a}$ .

(b.)

To determine

To Show: The given expressions are a parametrization for the line through the points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ .

Answer to Problem 48E

It has been shown that the given expressions are a parametrization for the line through the points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ .

Explanation of Solution

Given:

The expressions; $x=x_1+\left(x_2-x_1\right)t$ , $y=y_1+\left(y_2-y_1\right)t$ , where $-\infty <t<\infty$ .

Concept used:

The equation of the line through the points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is given as $\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}$ .

Calculation:

It is given that $x=x_1+\left(x_2-x_1\right)t$ , $y=y_1+\left(y_2-y_1\right)t$ , where $-\infty <t<\infty$ .

Simplifying,

  $x-x_1=\left(x_2-x_1\right)t$ ,

  $y-y_1=\left(y_2-y_1\right)t$

Dividing the second equation by the first equation,

  $\frac{y-y_1}{x-x_1}=\frac{\left(y_2-y_1\right)t}{\left(x_2-x_1\right)t}$

Simplifying,

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

As discussed, this is the equation of the line through the points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ .

Conclusion:

It has been shown that the given expressions are a parametrization for the line through the points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ .

(c.)

To determine

A parametrization for the line segment with the given end points.

Answer to Problem 48E

It has been determined that the parametrization for the line segment with the given end points; $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is; $x=x_1+\left(x_2-x_1\right)t$ , $y=y_1+\left(y_2-y_1\right)t$ , where $0\le t\le 1$ .

Explanation of Solution

Given:

The endpoints $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ .

Concept used:

The equation of the line through the points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is given as $\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}$ .

Calculation:

As discussed, the equation of the line through the points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is given as $\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}$ .

Simplifying the equation,

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

Now, let $\frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}=t$ .

Then the two equations obtained are:

  $\frac{y-y_1}{y_2-y_1}=t$ ,

  $\frac{x-x_1}{x_2-x_1}=t$

Simplifying,

  $y-y_1=\left(y_2-y_1\right)t$ ,

  $x-x_1=\left(x_2-x_1\right)t$

On further simplification,

  $y=y_1+\left(y_2-y_1\right)t$ ,

  $x=x_1+\left(x_2-x_1\right)t$

Now, put $\left(x,y\right)=\left(x_1,y_1\right)$ in the above equations to get,

  $y_1=y_1+\left(y_2-y_1\right)t$ ,

  $x_1=x_1+\left(x_2-x_1\right)t$

Simplifying,

  $\left(y_2-y_1\right)t=0$ ,

  $\left(x_2-x_1\right)t=0$

This gives $t=0$ .

Similarly, put $\left(x,y\right)=\left(x_2,y_2\right)$ in the obtained equations to get,

  $y_2=y_1+\left(y_2-y_1\right)t$ ,

  $x_2=x_1+\left(x_2-x_1\right)t$

Simplifying,

  $y_2-y_1=\left(y_2-y_1\right)t$ ,

  $x_2-x_1=\left(x_2-x_1\right)t$

This gives $t=1$ .

So, the required parametrization is; $x=x_1+\left(x_2-x_1\right)t$ , $y=y_1+\left(y_2-y_1\right)t$ , where $0\le t\le 1$ .

Conclusion:

It has been determined that the parametrization for the line segment with the given end points; $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is; $x=x_1+\left(x_2-x_1\right)t$ , $y=y_1+\left(y_2-y_1\right)t$ , where $0\le t\le 1$ .