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 = √(( 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=√((~~ 3- 2~~)^2 + (~~ 1- 3~~)^2) \\\\\\ ~\hfill EF=√(( 1)^2 + ( -2)^2) \implies \boxed{EF=√( 5 )}`

`F(\stackrel{x_1}{3}~,~\stackrel{y_1}{1})\qquad D(\stackrel{x_2}{-1}~,~\stackrel{y_2}{-1}) ~\hfill FD=√((~~ -1- 3~~)^2 + (~~ -1- 1~~)^2) \\\\\\ ~\hfill FD=√(( -4)^2 + ( -2)^2) \implies \boxed{FD=√( 20 )} \\\\\\ D(\stackrel{x_1}{-1}~,~\stackrel{y_1}{-1})\qquad E(\stackrel{x_2}{2}~,~\stackrel{y_2}{3}) ~\hfill DE=√((~~ 2- (-1)~~)^2 + (~~ 3- (-1)~~)^2) \\\\\\ ~\hfill DE=√(( 3)^2 + (4)^2) \implies DE=√( 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 }`