Chapter 1.1, Problem 34E

a .

To determine

To plot: the points $\left(2,10\right)$ and $\left(10,2\right)$

Answer to Problem 34E

Explanation of Solution

To plot the point $\left(2,10\right)$ , imagine a vertical line through 2 on the x -axis and a horizontal line through 10 on the y -axis. The intersection of these two lines is the point $\left(2,10\right)$ .

Similarly, to plot the point $\left(10,2\right)$ , imagine a vertical line through 10 on the x -axis and a horizontal line through 2 on the y -axis. The intersection of these two lines is the point $\left(10,2\right)$ .

The graph showing the points $\left(2,10\right)$ and $\left(10,2\right)$ is as given below:

b.

To determine

To find:the distance between the points $\left(2,10\right)$ and $\left(10,2\right)$ .

Answer to Problem 34E

  $d=11.31$

Explanation of Solution

Concept Used:

Distance Formula: Distance between the two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is given by:

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

Calculation:

In the above distance formula substitute $\left(x_1,y_1\right)=\left(2,10\right)$ and $\left(x_2,y_2\right)=\left(10,2\right)$ .

So, the distance between $\left(2,10\right)$ and $\left(10,2\right)$ is:

  $\begin{array}{c}d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(10-2\right)}^2+{\left(2-10\right)}^2}\\ =\sqrt{8^2+{\left(-8\right)}^2}\end{array}$

  $\begin{array}{c}=\sqrt{64+64}\\ =\sqrt{128}\\ =11.31\end{array}$

Conclusion:

The distance between $\left(2,10\right)$ and $\left(10,2\right)$ is 11.31 units.

c.

To determine

To find: the midpoint of the line segment joining the points $\left(2,10\right)$ and $\left(10,2\right)$ .

Answer to Problem 34E

  $\left(6,6\right)$

Explanation of Solution

Concept Used:

Midpoint Formula: Midpoint of the line segment joining the two points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is given by:

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

Calculation:

In the above midpoint formula substitute $\left(x_1,y_1\right)=\left(2,10\right)$ and $\left(x_2,y_2\right)=\left(10,2\right)$ .

So, the midpoint of the line segment joining $\left(2,10\right)$ and $\left(10,2\right)$ is:

  $\begin{array}{c}\text{Midpoint}=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\\ =\left(\frac{2+10}{2},\frac{10+2}{2}\right)\\ =\left(\frac{12}{2},\frac{12}{2}\right)\\ =\left(6,6\right)\end{array}$

Conclusion:

The midpoint of the line segment joining $\left(2,10\right)$ and $\left(10,2\right)$ is $\left(6,6\right)$ .