Find the perimeter of the triangle with the vertices at (2,0) (-1,6) (-5,-3)?

Answer 1

The distance between two points is given by the formula:

d = sqrt[(Δx)² + (Δy)²)

Where Δx is the difference is x coordinates and Δy is the difference in y coordinates.

You just have to find the lengths of the 3 sides and add them up.

(2,0) to (-1, 6):
Δx = 2 - (-1) = 3
Δy = 0 - 6 = -6
d1 = sqrt(3² + (-6)²)
d1 = sqrt(9 + 36)
d1 = sqrt(45)
d1 ≈ 6.71

(-1, 6) to (-5, -3):
Δx = (-1) - (-5) = 4
Δy = 6 - (-3) = 9
d2 = sqrt(4² + 9²)
d2 = sqrt(16 + 81)
d2 = sqrt(97)
d2 ≈ 9.85

(-5, -3) to (2,0):
Δx = (-5) - 2 = -7
Δy = (-3) - 0 = -3
d3 = sqrt((-7)² + (-3)²)
d3 = sqrt(49 + 9)
d3 = sqrt(58)
d3 ≈ 7.62

P = d1 + d2 + d3
P = 6.71 + 9.85 + 7.62
P ≈ 24.18

Answer 2

Use the Distance Formula to find the length of each side, then add the sides. Distance formula: Length of side = sqrt((x2 - x1)^2 + (y2 - y1)) The points you pick for X1,Y1 and X2,Y2 are arbitrary.

Answer 3

sqrt((2+1)^2+(0-6)^2)+
sqrt((-1+5)^2+(6+3)^2)+
sqrt((-5-2)^2+(-3-0)^2)

=24.17
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