Chapter 12, Problem 11PS

a.

To determine

To write: the distance d between the line and the point (3,1) as a function of m .

Answer to Problem 11PS

d= $\frac{\left|3m+3\right|}{\sqrt{m^2+1}}$ .

Explanation of Solution

Given information: A line with slope m passes through the point (0,4).

Formula used:

The distance d between a point $(x_1,y_1)$ and the line $Ax+By+C=0$ is given by;

  $d=\frac{\left|A{x}_1+B{y}_1+C\right|}{\sqrt{A^2+B^2}}.$

Calculation:

Given line has slope $m\Rightarrow y=mx+c.$As it pass through (0,4) $\Rightarrow 4=m\times 0+c\Rightarrow c=4.$

Thus line is $y=mx+4\Rightarrow mx-y+4=0.$

  $\begin{array}{c}hereA=m,B=-1andC=4,x_1=3,y_1=1.\\ d=\frac{\left|A{x}_1+B{y}_1+C\right|}{\sqrt{A^2+B^2}}=\frac{\left|m\times 3+(-1)\times 1+4\right|}{\sqrt{m^2+{(-1)}^2}}=\frac{\left|3m+3\right|}{\sqrt{m^2+1}}.\\ d=\frac{\left|3m+3\right|}{\sqrt{m^2+1}}.\end{array}$

b.

To determine

To graph: the function from part (a) using graphing utility.

Answer to Problem 11PS

Explanation of Solution

Given information: A line with slope m passes through the point (0,4).

Formula used:

The distance d between a point $(x_1,y_1)$ and the line $Ax+By+C=0$ is given by;

  $d=\frac{\left|A{x}_1+B{y}_1+C\right|}{\sqrt{A^2+B^2}}.$

Calculation:

d= $\frac{\left|3m+3\right|}{\sqrt{m^2+1}}$ .

The graph of the above function using graphing utility is given below.

  

Where x -axis represents m values and y -axis as d values.

c.

To determine

To Find : $\underset{m\to \infty }{\lim }d(m)and\underset{x\to -\infty }{\lim }d(m).$ Give a geometric interpretation of the results.

Answer to Problem 11PS

  $\underset{m\to \infty }{\lim }d(m)=3.$ .

  $\underset{m\to -\infty }{\lim }d(m)=3.$This indicates that the distance between the point and line approaches 3 as the slope approaches positive or negative.

Explanation of Solution

Given information: A line with slope m passes through the point (0,4).

Formula used:

The distance d between a point $(x_1,y_1)$ and the line $Ax+By+C=0$ is given by;

  $d=\frac{\left|A{x}_1+B{y}_1+C\right|}{\sqrt{A^2+B^2}}.$

Calculation:

d= $\frac{\left|3m+3\right|}{\sqrt{m^2+1}}$ .

  $\begin{array}{l}\underset{m\to \infty }{\lim }d(m)=\underset{m\to \infty }{\lim }\frac{\left|3m+3\right|}{\sqrt{m^2+1}}=3.\\ \underset{m\to \infty }{\lim }d(m)=3.\\ \underset{m\to -\infty }{\lim }d(m)=\underset{m\to -\infty }{\lim }\frac{\left|3m+3\right|}{\sqrt{m^2+1}}=3.\\ \underset{m\to -\infty }{\lim }d(m)=3.\end{array}$

Thus, this indicates that the distance between the point and line approaches 3 as the slope approaches positive or negative.