Respuesta :
[tex]\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&C&(~ -3 &,& 1~)
% (c,d)
&D&(~ 5 &,& 6~)
\end{array}~
% distance value
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
CD=\sqrt{[5-(-3)]^2+[6-1]^2}\implies CD=\sqrt{(5+3)^2+(6-1)^2}
\\\\\\
CD=\sqrt{8^2+5^2}\implies CD=\sqrt{64+25}\implies CD=\sqrt{89}[/tex]
[tex]\bf ~~~~~~~~~~~~\textit{middle point of 2 points }\\\\ \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &F&(~ -6 &,& 4~) % (c,d) &G&(~ 8 &,& -2~) \end{array}\qquad % coordinates of midpoint \left(\cfrac{ x_2 + x_1}{2}\quad ,\quad \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{8-6}{2}~~,~~\cfrac{-2+4}{2} \right)\implies \left( \cfrac{2}{2}~~,~~\cfrac{2}{2} \right)\implies (1,1)[/tex]
[tex]\bf \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &&(~ -2 &,& -3~) % (c,d) &&(~ 1 &,& 1~) \end{array} \\\\\\ % slope = m slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{1-(-3)}{1-(-2)}\implies \cfrac{1+3}{1+2}\implies \cfrac{4}{3}[/tex]
[tex]\bf ~~~~~~~~~~~~\textit{middle point of 2 points }\\\\ \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &F&(~ -6 &,& 4~) % (c,d) &G&(~ 8 &,& -2~) \end{array}\qquad % coordinates of midpoint \left(\cfrac{ x_2 + x_1}{2}\quad ,\quad \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{8-6}{2}~~,~~\cfrac{-2+4}{2} \right)\implies \left( \cfrac{2}{2}~~,~~\cfrac{2}{2} \right)\implies (1,1)[/tex]
[tex]\bf \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &&(~ -2 &,& -3~) % (c,d) &&(~ 1 &,& 1~) \end{array} \\\\\\ % slope = m slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{1-(-3)}{1-(-2)}\implies \cfrac{1+3}{1+2}\implies \cfrac{4}{3}[/tex]
