## College Algebra (11th Edition)

$\text{a) Distance: } \sqrt{34} \text{ units}\\\text{b) Midpoint: } \left( \dfrac{11}{2}, \dfrac{7}{2} \right)$
$\bf{\text{Solution Outline:}}$ Use the Distance Formula and the Midpoint Formula to find the distance and the midpoint of the given points $\left( 8,2 \right)$ and $\left( 3,5 \right).$ $\bf{\text{Solution Details:}}$ With the given points, then $x_1= 8 ,$ $x_2= 3 ,$ $y_1= 2 ,$ and $y_2= 5 .$ Using the Distance Formula which is given by $d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2} ,$ then \begin{array}{l}\require{cancel} d=\sqrt{(8-3)^2+(2-5)^2} \\\\ d=\sqrt{(5)^2+(-3)^2} \\\\ d=\sqrt{25+9} \\\\ d=\sqrt{34} .\end{array} Using $\left( \dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2} \right)$ or the Midpoint Formula, then the midpoint of the line segment with endpoints given above is \begin{array}{l}\require{cancel} \left( \dfrac{8+3}{2}, \dfrac{2+5}{2} \right) \\\\= \left( \dfrac{11}{2}, \dfrac{7}{2} \right) .\end{array} Hence, $\text{a) Distance: } \sqrt{34} \text{ units}\\\text{b) Midpoint: } \left( \dfrac{11}{2}, \dfrac{7}{2} \right) .$