Chapter 13.1, Problem 10CE

a.

To determine

To find the distance between the points

Answer to Problem 10CE

The value of d are $\sqrt{34}$

Explanation of Solution

The given point is

  $\left(0,0\right)\text{ and }\left(5,-3\right)$

Calculation:

The distance can be obtained as

Let, $\left(x_1,y_1\right)\text{ be }\left(0,0\right)\text{ and }\left(x_2,y_2\right)\text{ be }\left(5,-3\right)$

Use the distance formula between the two points

  $d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Put the values of $x_1=0,x_2=5,y_1=0\text{ and }{y}_2=-3$ on the above equation

  $\begin{array}{l}d=\sqrt{{\left(5-0\right)}^2+{\left(-3-0\right)}^2}\left[d=\text{distance}\right]\\ d=\sqrt{{\left(5\right)}^2+{\left(-3\right)}^2}\\ d=\sqrt{25+9}\\ d=\sqrt{34}\end{array}$

Hence,

The value of d are $\sqrt{34}$

b.

To determine

To find the distance between the points

Answer to Problem 10CE

The value of d are $8$

Explanation of Solution

The given point is

  $\left(3,-2\right)\text{ and }\left(-5,-2\right)$

Calculation:

The distance can be obtained as

Let, $\left(x_1,y_1\right)\text{ be }\left(3,-2\right)\text{ and }\left(x_2,y_2\right)\text{ be }\left(-5,-2\right)$

Use the distance formula between the two points

  $d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Put the values of $x_1=3,x_2=-5,y_1=-2\text{ and }{y}_2=-2$ on the above equation

  $\begin{array}{l}d=\sqrt{{\left(-5-3\right)}^2+{\left(-2-\left(-2\right)\right)}^2}\left[d=\text{distance}\right]\\ d=\sqrt{{\left(-8\right)}^2+{\left(-2+2\right)}^2}\\ d=\sqrt{{\left(-8\right)}^2+{\left(0\right)}^2}\\ d=\sqrt{64+0}\\ d=\sqrt{64}\\ d=8\end{array}$

Hence,

The value of d are $8$

c.

To determine

To find the distance between the points

Answer to Problem 10CE

The value of d are $\sqrt{98}$

Explanation of Solution

The given point is

  $\left(4,4\right)\text{ and }\left(-3,-3\right)$

Calculation:

The distance can be obtained as

Let, $\left(x_1,y_1\right)\text{ be }\left(4,4\right)\text{ and }\left(x_2,y_2\right)\text{ be }\left(-3,-3\right)$

Use the distance formula between the two points

  $d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Put the values of $x_1=4,x_2=-3,y_1=4\text{ and }{y}_2=-3$ on the above equation

  $\begin{array}{l}d=\sqrt{{\left(-3-4\right)}^2+{\left(-3-4\right)}^2}\left[d=\text{distance}\right]\\ d=\sqrt{{\left(-7\right)}^2+{\left(-7\right)}^2}\\ d=\sqrt{49+49}\\ d=\sqrt{98}\end{array}$

Hence,

The value of d are $\sqrt{98}$