Chapter 8, Problem 6T

Solution

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

The distance is $\sqrt{202}$ units and the midpoint is $\left(\frac{11}{2},\frac{7}{2}\right)$ .

Given information:

Given points are $\left(1,9\right)\text{ and }\left(10,-2\right)$ .

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 $\left(1,9\right)\text{ and }\left(10,-2\right)$ , 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 $\left(1,9\right)\text{ and }\left(10,-2\right)$ , thus the distance is calculated as:

  $\begin{array}{c}\sqrt{{\left(10-1\right)}^2+{\left(-2-9\right)}^2}=\sqrt{9^2+{\left(-11\right)}^2}\\ =\sqrt{81+121}\\ =\sqrt{202}\text{units}\end{array}$

Now, to find the midpoint of a line segment joining the points $\left(1,9\right)\text{ and }\left(10,-2\right)$ use the formula

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

As,

  $\begin{array}{c}\left(\frac{x_2+x_1}{2},\frac{y{2}_{}+y{1}_{}}{2}\right)=\left(\frac{10+1}{2},\frac{-2+9}{2}\right)\\ =\left(\frac{11}{2},\frac{7}{2}\right)\end{array}$

Thus, the distance between the given two points is $\sqrt{202}$ units and the midpoint of the line segment joining the given points is $\left(\frac{11}{2},\frac{7}{2}\right)$ .