Question

For the points (-2,-7) and (8,-11) (a) Find the exact distance between the points. (b) Find the midpoint of the line segment whose endpoints are the given points. Part 1 of 5 (a) Find the exact distance between the points. Label the points. Note that the choice for (x1,y1) and (x2,y2) will not affect the outcome. (-2,-7) and (8,-11) (x1,y1) and (x2,y2) Apply the distance formula. d=(x2-x1)^2+(y2-y1)^2

Answer 1

`~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-2}~,~\stackrel{y_1}{-7})\qquad (\stackrel{x_2}{8}~,~\stackrel{y_2}{-11})\qquad \qquad d = √(( x_2- x_1)^2 + ( y_2- y_1)^2) \\\\\\ d=√((~~8 - (-2)~~)^2 + (~~-11 - (-7)~~)^2)\implies d=√((8 +2)^2 + (-11 +7)^2) \\\\\\ d=√( (10)^2 + (-4)^2) \implies d=√( 100 + 16)\implies d=√( 116 )\implies d=2√(29) \\\\[-0.35em] ~\dotfill`

`~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{-2}~,~\stackrel{y_1}{-7})\qquad (\stackrel{x_2}{8}~,~\stackrel{y_2}{-11}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left(\cfrac{ 8 -2}{2}~~~ ,~~~ \cfrac{ -11 -7}{2} \right) \implies \left(\cfrac{ 6 }{2}~~~ ,~~~ \cfrac{ -18 }{2} \right)\implies (3~~,~-9)`