Chapter 8.5, Problem 5Q

Solution

The distance between given two points, and the midpoint of the line segment joining these points.

The distance is $\sqrt{194}$ units and the midpoint is $\left(-\frac{7}{2},\frac{3}{2}\right)$ .

Given information:

Given points are $\left(-6,-5\right)\text{ and }\left(-1,8\right)$ .

Formula used:

• The distance between two points, $\left(x_1,y{1}_{}\right),\text{ and }\left(x_2,y{2}_{}\right)$ , is given by:

  $\sqrt{{\left(x_2-x_1\right)}^2+{\left(y{2}_{}-y{1}_{}\right)}^2}$

• The midpoint of a line segment joining the points $\left(x_1,y{1}_{}\right),\text{ and }\left(x_2,y{2}_{}\right)$ , is given by:

  $\left(\frac{x_2+x_1}{2},\frac{y{2}_{}+y{1}_{}}{2}\right)$

Calculation:

To calculate the distance between $\left(-6,-5\right)\text{ and }\left(-1,8\right)$ , use the formula for the distance between two points $\left(x_1,y{1}_{}\right),\text{ and }\left(x_2,y{2}_{}\right)$ that is

  $\sqrt{{\left(x_2-x_1\right)}^2+{\left(y{2}_{}-y{1}_{}\right)}^2}$

Here, points are $\left(-6,-5\right)\text{ and }\left(-1,8\right)$ , thus the distance is calculated as:

  $\begin{array}{c}\sqrt{{\left(-1-\left(-6\right)\right)}^2+{\left(8-\left(-5\right)\right)}^2}=\sqrt{{\left(-1+6\right)}^2+{\left(8+5\right)}^2}\\ =\sqrt{{\left(5\right)}^2+{\left(13\right)}^2}\\ =\sqrt{25+169}\\ \text{=}\sqrt{194}\text{ units}\end{array}$

Now, to find the midpoint of a line segment joining the points $\left(-6,-5\right)\text{ and }\left(-1,8\right)$ use the formula

  $\left(\frac{x_2+x_1}{2},\frac{y{2}_{}+y{1}_{}}{2}\right)$

As,

  $\begin{array}{c}\left(\frac{x_2+x_1}{2},\frac{y{2}_{}+y{1}_{}}{2}\right)=\left(\frac{-1+\left(-6\right)}{2},\frac{8+\left(-5\right)}{2}\right)\\ =\left(\frac{-1-6}{2},\frac{8-5}{2}\right)\\ =\left(-\frac{7}{2},\frac{3}{2}\right)\end{array}$

Thus, the distance between the given two points is $\sqrt{194}$ units and the midpoint of the line segment joining the given points is $\left(-\frac{7}{2},\frac{3}{2}\right)$ .