Chapter 5.5, Problem 131E

(a)

To determine

Tograph: The points $\left(5,2\right)$ and $\left(-1,4\right)$ .

Explanation of Solution

Given information:

The given points are $\left(5,2\right)$ and $\left(-1,4\right)$ .

Graph:

  

Interpretation:

To plot a point on a coordinate plane, the value of $x$ represents the position along the x -axis and the value of $y$ represents the position along the y -axis.

To plot point $\left(5,2\right)$ , move 5 units along the x -axis and 2 units up.

To plot point $\left(-1,4\right)$ , move 1 unit along the negativex -axis and 4 units up.

Therefore, the graph is plotted above.

(b)

To determine

To find:The distance between the points $\left(5,2\right)$ and $\left(-1,4\right)$ .

Answer to Problem 131E

The distance between the points $\left(5,2\right)$ and $\left(-1,4\right)$ is $2\sqrt{10}$ .

Explanation of Solution

Given information:

The given points are $\left(5,2\right)$ and $\left(-1,4\right)$ .

Calculation:

The distance between two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ can be calculated by the distance formula.

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

Let the point $\left(x_1,y_1\right)$ is $\left(5,2\right)$ and $\left(-1,4\right)$ is $\left(x_2,y_2\right)$ .

Substitute the values in the above equation to calculate the distance.

  $\begin{array}{c}\text{Distance}=\sqrt{{\left(\left(-1\right)-5\right)}^2+{\left(4-2\right)}^2}\\ =\sqrt{{\left(-6\right)}^2+2^2}\\ =\sqrt{36+4}\\ =2\sqrt{10}\end{array}$

Therefore, the distance between the points $\left(5,2\right)$ and $\left(-1,4\right)$ is $2\sqrt{10}$ .

(c)

To determine

To find:The midpoint of the line segment connecting the points $\left(5,2\right)$ and $\left(-1,4\right)$ .

Answer to Problem 131E

The midpoint of the line segmentis $\left(2,3\right)$ .

Explanation of Solution

Given information:

The given points are $\left(5,2\right)$ and $\left(-1,4\right)$ .

Calculation:

The midpoint of the line segment connect the points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ can be calculated by the midpoint formula.

  $\text{Midpoint}=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$

Let the point $\left(x_1,y_1\right)$ is $\left(5,2\right)$ and $\left(-1,4\right)$ is $\left(x_2,y_2\right)$ .

Substitute the values in the above equation to calculate the distance.

  $\begin{array}{c}\text{Midpoint}=\left(\frac{5+\left(-1\right)}{2},\frac{2+4}{2}\right)\\ =\left(\frac{4}{2},\frac{6}{2}\right)\\ =\left(2,3\right)\end{array}$

Therefore, the midpoint of the line segment is $\left(2,3\right)$ .