Chapter 1.8, Problem 14E

  $\left(a\right)$

To determine

To graph: The points $\left(-2,5\right)\text{ and }\left(10,0\right)$ in a coordinate plane.

Explanation of Solution

Given information:

The points, $\left(-2,5\right)\text{ and }\left(10,0\right)$ .

Graph:

The plot the points $\left(-2,5\right)\text{ and }\left(10,0\right)$ in the coordinate plane.

Construct a Cartesian plane with vertical axis as y-axis and horizontal axis as x-axis.

Marks the numbers on both the axes.

Plot the points $\left(-2,5\right)\text{ and }\left(10,0\right)$ .

  

Interpretation:

The Cartesian plane is divided into four quadrant. Numbers on right of zero are positive, on left of zero are negative. Numbers above zero on vertical line are positive and numbers below 0 are negative. Each pair of point in the coordinate plane is represented as an ordered pair $\left(x,y\right)$ where first element is the x-coordinate and second element is the y-coordinate. The point $\left(x,y\right)$ denote it is x units on x-axis and y units on y −axis and marked accordingly.

  $\left(b\right)$

To determine

To calculate: The distance between the pair of points $\left(-2,5\right)\text{ and }\left(10,0\right)$ .

Answer to Problem 14E

The distance between the pair of points $\left(-2,5\right)\text{ and }\left(10,0\right)$ is $13$ units.

Explanation of Solution

Given information:

The points $\left(-2,5\right)\text{ and }\left(10,0\right)$ .

Formula used:

Distance $d\left(A,B\right)$ between two points $A=\left(x_1,y_1\right)\text{ and }B=\left(x_2,y_2\right)$ in the Cartesian plane is denoted by $d\left(A,B\right)=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$ .

Calculation:

Consider the points $\left(-2,5\right)\text{ and }\left(10,0\right)$ .

Recall that the distance $d\left(A,B\right)$ between two points $A=\left(x_1,y_1\right)\text{ and }B=\left(x_2,y_2\right)$ in the Cartesian plane is denoted by $d\left(A,B\right)=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$ .

Evaluate the distance between $\left(-2,5\right)\text{ and }\left(10,0\right)$ .

  $\begin{array}{c}d\left(A,B\right)=\sqrt{{\left(10-\left(-2\right)\right)}^2+{\left(0-5\right)}^2}\\ =\sqrt{144+25}\\ =\sqrt{169}\\ =13\end{array}$

Thus, the distance between the pair of points is $13$ units.

  $\left(c\right)$

To determine

To calculate: The midpoint of the segment that joins the pair of points $\left(-2,5\right)\text{ and }\left(10,0\right)$ .

Answer to Problem 14E

The midpoint of the segment that joins the pair of points $\left(-2,5\right)\text{ and }\left(10,0\right)$ is $\left(4,\frac{5}{2}\right)$ .

Explanation of Solution

Given information:

The points are $\left(-2,5\right)\text{ and }\left(10,0\right)$ .

Formula used:

Midpoint of the segment that joins two points $A=\left(x_1,y_1\right)\text{ and }B=\left(x_2,y_2\right)$ in the Cartesian plane is denoted by $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$ .

Calculation:

Consider the points $\left(-2,5\right)\text{ and }\left(10,0\right)$ .

Recall that the midpoint of the segment that joins two points $A=\left(x_1,y_1\right)\text{ and }B=\left(x_2,y_2\right)$ in the Cartesian plane is denoted by $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$ .

Evaluate the midpoint of the segment that joins $\left(-2,5\right)\text{ and }\left(10,0\right)$ .

  $\begin{array}{c}\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)=\left(\frac{-2+10}{2},\frac{5+0}{2}\right)\\ =\left(\frac{8}{2},\frac{5}{2}\right)\\ =\left(4,\frac{5}{2}\right)\end{array}$

Thus, the midpoint of the segment that joins the pair of points is $\left(4,\frac{5}{2}\right)$ .