Question

a civil engineer has laid a graph down over the stretch of a new road system. a circular rotary has a diamater with endpoints (-3,5) and (5,11) what are the center and raduis of the rotary

Answer 1

$\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{-3}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{5}~,~\stackrel{y_2}{11}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{5-3}{2}~~,~~\cfrac{11+5}{2} \right)\implies \left(\cfrac{2}{2}~,~ \cfrac{16}{2}\right)\implies \stackrel{center}{(1,8)} \\\\[-0.35em] ~\dotfill$

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ \stackrel{endpoint}{(\stackrel{x_1}{-3}~,~\stackrel{y_1}{5})}\qquad \stackrel{center}{(\stackrel{x_2}{1}~,~\stackrel{y_2}{8})}\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{radius}{r}=\sqrt{[1-(-3)]^2+[8-5]^2}\implies r=\sqrt{(1+3)^2+(8-5)^2} \\\\\\ r=\sqrt{16+9}\implies r=\sqrt{25}\implies r=5$

Answer 2

Do the midpoint formula. Which will give you (1,8)