Question

I’m doing my geometry homework and don’t remember the formula or way to solve the lengths of a triangle side with graph points. How do I solve both parts of #1?

Answer 1

$~\hfill \stackrel{\textit{\large distance between 2 points}}{d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}}~\hfill~ \\\\[-0.35em] ~\dotfill\\\\ E(\stackrel{x_1}{2}~,~\stackrel{y_1}{3})\qquad F(\stackrel{x_2}{3}~,~\stackrel{y_2}{1}) ~\hfill EF=\sqrt{(~~ 3- 2~~)^2 + (~~ 1- 3~~)^2} \\\\\\ ~\hfill EF=\sqrt{( 1)^2 + ( -2)^2} \implies \boxed{EF=\sqrt{ 5 }}$

$F(\stackrel{x_1}{3}~,~\stackrel{y_1}{1})\qquad D(\stackrel{x_2}{-1}~,~\stackrel{y_2}{-1}) ~\hfill FD=\sqrt{(~~ -1- 3~~)^2 + (~~ -1- 1~~)^2} \\\\\\ ~\hfill FD=\sqrt{( -4)^2 + ( -2)^2} \implies \boxed{FD=\sqrt{ 20 }} \\\\\\ D(\stackrel{x_1}{-1}~,~\stackrel{y_1}{-1})\qquad E(\stackrel{x_2}{2}~,~\stackrel{y_2}{3}) ~\hfill DE=\sqrt{(~~ 2- (-1)~~)^2 + (~~ 3- (-1)~~)^2} \\\\\\ ~\hfill DE=\sqrt{( 3)^2 + (4)^2} \implies DE=\sqrt{ 25 }\implies \boxed{DE=5} \\\\[-0.35em] \rule{34em}{0.25pt}$

$E(\stackrel{x_1}{2}~,~\stackrel{y_1}{3})\qquad F(\stackrel{x_2}{3}~,~\stackrel{y_2}{1}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{1}-\stackrel{y1}{3}}}{\underset{run} {\underset{x_2}{3}-\underset{x_1}{2}}} \implies \cfrac{ -2 }{ 1 } \implies - 2 \\\\[-0.35em] ~\dotfill$

$F(\stackrel{x_1}{3}~,~\stackrel{y_1}{1})\qquad D(\stackrel{x_2}{-1}~,~\stackrel{y_2}{-1}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{-1}-\stackrel{y1}{1}}}{\underset{run} {\underset{x_2}{-1}-\underset{x_1}{3}}} \implies \cfrac{ -2 }{ -4 } \implies \cfrac{1 }{ 2 } \\\\[-0.35em] ~\dotfill$

$D(\stackrel{x_1}{-1}~,~\stackrel{y_1}{-1})\qquad E(\stackrel{x_2}{2}~,~\stackrel{y_2}{3}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{3}-\stackrel{y1}{(-1)}}}{\underset{run} {\underset{x_2}{2}-\underset{x_1}{(-1)}}} \implies \cfrac{3 +1}{2 +1} \implies \cfrac{4 }{ 3 }$