Chapter 9.5, Problem 1PSA

a.

To determine

To calculate: The distance between points $(4,0)$ and $(6,0)$ .

Answer to Problem 1PSA

The distance between points $(4,0)$ and $(6,0)$ is 2.

Explanation of Solution

Given information:

Points are $(4,0)$ and $(6,0)$ .

Formula used:

The distance between two points by using the distance formula, which is an application of the Pythagoreantheorem.

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

Calculation:

Distance:

  $\begin{array}{l}D=\sqrt{{\left(6-4\right)}^2+{\left(0-0\right)}^2}\\ D=\sqrt{4+0}\\ D=\sqrt{4}\\ D=2\end{array}$

b.

To determine

To find: The distance between points $(2,3)$ and $(2,-1)$ .

Answer to Problem 1PSA

The distance between points $(2,3)$ and $(2,-1)$ is 4.

Explanation of Solution

Given information:

Points are $(2,3)$ and $(2,-1)$ .

Formula used:

The distance between two points by using the distance formula, which is an application of the Pythagoreantheorem.

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

Calculation:

Distance:

  $\begin{array}{l}D=\sqrt{{\left(2-2\right)}^2+{\left(3-(-1)\right)}^2}\\ D=\sqrt{0+16}\\ D=\sqrt{16}\\ D=4\end{array}$

c.

To determine

To calculate: The distance between points $(4,1)$ and $(7,5)$ .

Answer to Problem 1PSA

The distance between points $(4,1)$ and $(7,5)$ is 5.

Explanation of Solution

Given information:

Points are $(4,1)$ and $(7,5)$ .

Formula used:

The distance between two points by using the distance formula, which is an application of the Pythagoreantheorem.

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

Calculation:

Distance:

  $\begin{array}{l}D=\sqrt{{\left(7-4\right)}^2+{\left(5-1\right)}^2}\\ D=\sqrt{9+16}\\ D=\sqrt{25}\\ D=5\end{array}$

d.

To determine

To find: The distance between points $(-2,-4)$ and $(-8,4)$ .

Answer to Problem 1PSA

The distance between points $(-2,-4)$ and $(-8,4)$ is 10.

Explanation of Solution

Given information:

Points are $(-2,-4)$ and $(-8,4)$ .

Formula used:

The distance between two points by using the distance formula, which is an application of the Pythagoreantheorem.

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

Calculation:

Distance:

  $\begin{array}{l}D=\sqrt{{\left(4-(-4)\right)}^2+{\left(-8-(-2)\right)}^2}\\ D=\sqrt{8^2+6^2}\\ D=\sqrt{100}\\ D=10\end{array}$

e.

To determine

To calculate:The distance between points $(0,0)$ and $(2,5)$ .

Answer to Problem 1PSA

The distance between points $(0,0)$ and $(2,5)$ is $\sqrt{29}$ .

Explanation of Solution

Given information:

Points are $(0,0)$ and $(2,5)$ .

Formula used:

The distance between two points by using the distance formula, which is an application of the Pythagoreantheorem.

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

Calculation:

Distance:

  $\begin{array}{l}D=\sqrt{{\left(2-0\right)}^2+{\left(5-0\right)}^2}\\ D=\sqrt{4+25}\\ D=\sqrt{29}\end{array}$

f.

To determine

To find: The distance between points $(2,1)$ and $(6,3)$ .

Answer to Problem 1PSA

The distance between points $(2,1)$ and $(6,3)$ is $2\sqrt{5}$ .

Explanation of Solution

Given information:

Points are $(2,1)$ and $(6,3)$ .

Formula used:

The distance between two points by using the distance formula, which is an application of the Pythagoreantheorem.

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

Calculation:

Distance:

  $\begin{array}{l}D=\sqrt{{\left(6-2\right)}^2+{\left(3-1\right)}^2}\\ D=\sqrt{16+4}\\ D=\sqrt{20}\\ D=2\sqrt{5}\end{array}$