Chapter 3, Problem 1CLT

a .

To determine

To plot: the points $\left(-2,5\right)$ and $\left(3,-1\right)$

Answer to Problem 1CLT

  

Explanation of Solution

To plot the point $\left(-2,5\right)$ , imagine a vertical line through -2 on the x -axis and a horizontal line through 5 on the y -axis. The intersection of these two lines is the point $\left(-2,5\right)$ .

Similarly, to plot the point $\left(3,-1\right)$ , imagine a vertical line through 3 on the x -axis and a horizontal line through -1 on the y -axis. The intersection of these two lines is the point $\left(3,-1\right)$ .

The graph showing the points $\left(-2,5\right)$ and $\left(3,-1\right)$ is as given below:

b.

To determine

To find: the distance between the points $\left(-2,5\right)$ and $\left(3,-1\right)$ .

Answer to Problem 1CLT

  $d=\sqrt{61}$

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 $\left(x_1,y_1\right)=\left(-2,5\right)$ and $\left(x_2,y_2\right)=\left(3,-1\right)$ .

So, the distance between $\left(-2,5\right)$ and $\left(3,-1\right)$ is:

  $\begin{array}{c}d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(3-\left(-2\right)\right)}^2+{\left(-1-5\right)}^2}\\ =\sqrt{{\left(3+2\right)}^2+{\left(-1-5\right)}^2}\end{array}$

  $\begin{array}{c}=\sqrt{5^2+{\left(-6\right)}^2}\\ =\sqrt{25+36}\\ =\sqrt{61}\end{array}$

Conclusion:

The distance between $\left(-2,5\right)$ and $\left(3,-1\right)$ is $\sqrt{61}$ units.

c.

To determine

To find: the midpoint of the line segment joining the points $\left(-2,5\right)$ and $\left(3,-1\right)$ .

Answer to Problem 1CLT

  $\left(\frac{1}{2},2\right)$

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 $\left(x_1,y_1\right)=\left(-2,5\right)$ and $\left(x_2,y_2\right)=\left(3,-1\right)$ .

So, the midpoint of the line segment joining $\left(-2,5\right)$ and $\left(3,-1\right)$ is:

  $\begin{array}{c}\text{Midpoint}=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)\\ =\left(\frac{-2+3}{2},\frac{5+\left(-1\right)}{2}\right)\\ =\left(\frac{1}{2},\frac{4}{2}\right)\\ =\left(\frac{1}{2},2\right)\end{array}$

Conclusion:

The midpoint of the line segment joining $\left(-2,5\right)$ and $\left(3,-1\right)$ is $\left(\frac{1}{2},2\right)$ .