How do you find the perimeter of a triangle on a coordinate plane whose points are A (-3, 6), B (-3, 2), C (3, 2)?

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Answer 1 · 2015-06-07T00:52:26

Be a triangle formed with the points $A [- 3; 6]$ $A[-3;6]$ , $B [- 3; 2]$ $B[-3;2]$ and $C [3; 2]$ $C[3;2]$ .

The perimeter of this triangle is

$Pi = AB+BC+AC$

In a plane, the distance between two points M and N is given by

$d_(MN) = sqrt((x_M-x_N)^2+(y_M-y_N)^2)$

Therefore

$AB = sqrt((-3-(-3))^2+(6-2)^2) = sqrt(0+4^2) = 4$

$BC = sqrt((-3-3)^2+(2-2)^2) = sqrt((-6)^2+0) = 6$

$AC = sqrt((-3-3)^2+(6-2)^2) = sqrt((-6)^2+4^2) = sqrt52$

$rarr Pi = 4+6+sqrt52 = 10+sqrt52 = 10+2sqrt13$