Find the perimeter of the triangle with the vertices at (1, -2), (-3, 2), and (-2, -4).?

Let:

V1 =(x1 , y1) = (1, -2 )

V2 =(x2 , y2) = (- 3, 2)

V3 =(x3 , y3) = (-2, -4)

Distance between two vertices is = Distance between two points

d^2 = (x2 - x1)^2 + (y2 - y1)^2

d = sqrt[(x2 - x1)^2 + (y2 - y1)^2]

Distance between vertices V1 and V2:

d1 = sqrt[(x2 - x1)^2 + (y2 - y1)^2]

d1 = sqrt[( -3 - 1)^2 + {(2 - (-2)}^2 ]

d1 = sqrt[( -4)^2 + (2+ 2^2]

d1 = sqrt(16 + 16)= sqrt(32)

d1 = 5.66

Distance between vertices V3 and V2:

d2 = sqrt[ {(-2 - (- 3)}^2 + ( - 4 - 2)^2]

d2= sqrt( 1 + 36)= sqrt(37)

d2= 6.08...

Distance between vertices V3 and V1:

d3 = sqrtt[(- 2 - 1)^2 +{ (- 4 - (-2)}^2 ]

d3= sqrt(9 + 4)

d3= sqrt(13) = 3.61

P = d1 + d2 + d3

P=5.66 + 6.08 + 3.61

P = 15.35 .....<===Answer for the peremeter

4

You can solve for the length of each side by

d = sqrt((y2-y1)^2 + (x2-x1)^2)