Question

In Exercises 6-1

6.
Find the perimeter of ABC.

A(-4,3)
B(-1,5)
C(2,3)

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\\\\ A(\stackrel{x_1}{-4}~,~\stackrel{y_1}{3})\qquad B(\stackrel{x_2}{-1}~,~\stackrel{y_2}{5}) ~\hfill AB=√((~~ -1- (-4)~~)^2 + (~~ 5- 3~~)^2) \\\\\\ ~\hfill AB=√(( 3)^2 + ( 2)^2)\implies \boxed{AB=√(13)}`

`B(\stackrel{x_1}{-1}~,~\stackrel{y_1}{5})\qquad C(\stackrel{x_2}{2}~,~\stackrel{y_2}{3}) ~\hfill BC=√((~~ 2- (-1)~~)^2 + (~~ 3- 5~~)^2) \\\\\\ ~\hfill BC=√(( 3)^2 + ( -2)^2)\implies \boxed{BC=√(13)} \\\\\\ C(\stackrel{x_1}{2}~,~\stackrel{y_1}{3})\qquad A(\stackrel{x_2}{-4}~,~\stackrel{y_2}{3}) ~\hfill CA=√((~~ -4- 2~~)^2 + (~~ 3- 3~~)^2) \\\\\\ ~\hfill CA=√(( -6)^2 + (0)^2)\implies \boxed{CA=6}`

`~\dotfill\\\\ \stackrel{\textit{\LARGE Perimeter}}{√(13)~~ + ~~√(13)~~ + ~~6 ~~ \approx ~~ \text{\LARGE 13.21}}`