MHB Are the coordinates of the vertices of this triangle all integers?

AI Thread Summary
The discussion revolves around determining the vertices of a triangle defined by three vector equations. The calculated vertices are (-3,2), (7,0), and (3,5), but the resulting perimeter is not an integer, raising concerns about the calculations. The perimeter is expressed in terms of square roots, leading to speculation that the original problem may have intended for the vertices to be integers. Participants agree on the Cartesian conversions of the lines and confirm the perimeter calculations. The conversation concludes with the suggestion that the problem might have been misinterpreted regarding the integer requirement for the vertices.
chucktingle
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L1 [x,y]=[2,1]+r[-5,1]
L2 [x,y]=[1,4]+s[2,1]
L3 [x,y]=[3,5]+t[4,-5]
These three lines are sides of a triangle
find: 1)the perimeter of the triangle
2) The largest angle
3) the centroid of the triangle

so I converted the vector equations into parametric, and then made two of the x parametric equations to equal each other to find the vertices.This gave me to vertices (-3,2), (7,0) and (3,5). The problem is the perimeter I get from those vertices is not an integer, I was told it would have no decimals. Am I going about the problem incorrectly? Once I have the vertices I can easily find the angle with cosine law, and use the centroid formula for the centroid, I am just not sure about the vertices I got.
 
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I think I would convert to Cartesian coordinates:

$$L1\implies x+5y=7$$

$$L2\implies x-2y=-7$$

$$L3\implies 5x+4y=35$$

View attachment 6567

I agree with the vertices you found. And so the perimeter $P$ is:

$$P=\sqrt{6^2+3^2}+\sqrt{4^2+5^2}+\sqrt{10^2+2^2}=3\sqrt{5}+\sqrt{41}+2\sqrt{26}$$

Is this what you have?
 

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MarkFL said:
I think I would convert to Cartesian coordinates:

$$L1\implies x+5y=7$$

$$L2\implies x-2y=-7$$

$$L3\implies 5x+4y=35$$
I agree with the vertices you found. And so the perimeter $P$ is:

$$P=\sqrt{6^2+3^2}+\sqrt{4^2+5^2}+\sqrt{10^2+2^2}=3\sqrt{5}+\sqrt{41}+2\sqrt{26}$$

Is this what you have?

Thanks for the reply, yep that's what I got. Maybe they rounded for the answer?
 
chucktingle said:
Thanks for the reply, yep that's what I got. Maybe they rounded for the answer?

I'm thinking that perhaps what was intended was that the coordinates of the 3 vertices would all be integers. :)
 
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