Chapter 8, Problem 1T

Solution

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

The distance is $2\sqrt{17}$ units and the midpoint is $\left(3,4\right)$ .

Given information:

Given points are $\left(-1,5\right)\text{ and }\left(7,3\right)$ .

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 $\left(-1,5\right)\text{ and }\left(7,3\right)$ , 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 $\left(-1,5\right)\text{ and }\left(7,3\right)$ , thus the distance is calculated as:

  $\begin{array}{c}\sqrt{{\left(7-\left(-1\right)\right)}^2+{\left(3-5\right)}^2}=\sqrt{{\left(7+1\right)}^2+{\left(-2\right)}^2}\\ =\sqrt{{\left(8\right)}^2+4}\\ =\sqrt{64+4}\\ =\sqrt{68}\\ =2\sqrt{17}\text{ units}\end{array}$

Now, to find the midpoint of a line segment joining the points $\left(-1,5\right)\text{ and }\left(7,3\right)$ use the formula

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

As,

  $\begin{array}{c}\left(\frac{x_2+x_1}{2},\frac{y{2}_{}+y{1}_{}}{2}\right)=\left(\frac{7+\left(-1\right)}{2},\frac{3+5}{2}\right)\\ =\left(\frac{7-1}{2},\frac{8}{2}\right)\\ =\left(\frac{6}{2},4\right)\\ =\left(3,4\right)\end{array}$

Thus, the distance between the given two points is $2\sqrt{17}$ units and the midpoint of the line segment joining the given points is $\left(3,4\right)$ .