Chapter 13, Problem 5RP

(a)

To determine

To find: The value of $PQ$ .

Answer to Problem 5RP

The value of $PQ=10$ .

Explanation of Solution

Given:

The point $P=\left(4,-2\right)$ and $Q=\left(10,6\right)$ .

Calculation:

The distance between the two points in the coordinate plane is of the form,

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

Then,

  $\begin{array}{c}PQ=\sqrt{{\left(10-4\right)}^2+{\left(6-\left(-2\right)\right)}^2}\\ =10\end{array}$

(b)

To determine

To find: The midpoint of PQ.

Answer to Problem 5RP

The midpoint of PQ is $\left(7,2\right)$ .

Explanation of Solution

Given:

The point $P=\left(4,-2\right)$ and $Q=\left(10,6\right)$ .

Calculation:

The formula to determine the mid-point of the line is of the form,

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

Then,

  $\begin{array}{c}M=\left(\frac{4+10}{2},\frac{-2+6}{2}\right)\\ =\left(7,2\right)\end{array}$

(c)

To determine

To find: The slope of the PQ.

Answer to Problem 5RP

The slope of the PQ is $1\frac{1}{3}$ .

Explanation of Solution

Given:

The point $P=\left(4,-2\right)$ and $Q=\left(10,6\right)$ .

Calculation:

The formula to determine the slope of the line is of the form,

  $m=\frac{y_2-y_1}{x_2-x_1}$

Then,

  $\begin{array}{c}m=\frac{6-\left(-1\right)}{10-4}\\ =\frac{4}{3}\\ =1\frac{1}{3}\end{array}$