SOLUTION: how do i find the perimeter of the triangle with vertices at three points (0,1),(2,1), and (0,3)? I know that P= a+b+c, so will the next step be the sq root (X1-X2)+(Y1-Y2)+(Z1-

Algebra ->  Points-lines-and-rays -> SOLUTION: how do i find the perimeter of the triangle with vertices at three points (0,1),(2,1), and (0,3)? I know that P= a+b+c, so will the next step be the sq root (X1-X2)+(Y1-Y2)+(Z1-      Log On


   



Question 212943This question is from textbook Geometry connections
: how do i find the perimeter of the triangle with vertices at three points (0,1),(2,1), and (0,3)?
I know that P= a+b+c, so will the next step be the sq root (X1-X2)+(Y1-Y2)+(Z1-Z2)?
This question is from textbook Geometry connections

Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
To get the perimeter you have to add the length of each side.
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how do i find the perimeter of the triangle with vertices at three points (0,1),(2,1), and (0,3)?
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summing up a + b + c if each represents a side of the triangle is good.
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I know that P= a+b+c, so will the next step be the sq root (X1-X2)+(Y1-Y2)+(Z1-Z2)?
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I think you have the right idea.
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the length of a line is equal to sqrt%28%28x2-x1%29%5E2+%2B+%28y2-y1%29%5E2%29
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Your points are:
A = (0,1)
B = (2,1)
C = (0,3)
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Side a would be opposite angle A would be side BC
Side b would be opposite angle B would be side AC
Side c would be opposited angle C would be side AB
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I'll do one side.
You can do the rest.
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Side a is line BC.
Points on line BC are:
(x1,y1) = point B = (2,1)
(x2,y2) = Point C = (0,3)
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(x2-x1) = 0-2 = -2
(y2-y1) = 3-1 = 2
length of line BC = sqrt%28%28-2%29%5E2+%2B+%282%29%5E2%29 = sqrt%284%2B4%29 = sqrt%288%29 = 2.828427125
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