Find 3rd Point of Right Angled Triangle

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SUMMARY

The discussion focuses on determining the coordinates of the third point in a right-angled triangle given the lengths of its sides (13, 9, and approximately 16) and the coordinates of two points (273, 541) and (273, 525). Participants suggest using vector decomposition, slopes, and line equations to solve the problem. The conversation emphasizes the need for a solid understanding of vector geometry and algebra to programmatically calculate the third point for any triangle configuration.

PREREQUISITES
  • Understanding of right-angled triangles and their properties
  • Familiarity with vector decomposition and components
  • Knowledge of slopes and line equations in coordinate geometry
  • Basic programming skills for implementing mathematical solutions
NEXT STEPS
  • Study vector decomposition and how to calculate vector components
  • Learn about slopes and their relationship in coordinate geometry
  • Explore the distance formula and its application in triangle geometry
  • Research algorithms for calculating triangle vertices programmatically
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Mathematicians, programmers, and students working on geometry problems, particularly those involving right-angled triangles and vector mathematics.

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OK I have right angled triangle, I no the length of the 3 sides. and the position
of two of the points but i want to be able to figure where the 3rd point is. I
attached a rough sort of diagram so u no wot I'm talkin about. I 'm sure there
is enuf info there to find the point but i just can't get my head round it

Lenght of sides 13, 9 and SQROOT of 13^2 + 9^2 = 16(roughly)
Points = Bottom -> (273,541)
Top -> (273,525)


http://studentweb.itsligo.ie/business/kcomp_b4/s00001829/angles.jpg
 
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Do you know how to decompose vectors into components?

- Warren
 
I don't know wot that is??
Could u explain?
 
A vector is a directed line segment -- a distance and a direction.

The components of a vector \vec u on a plane are (l \cos \theta, l \sin \theta) where l is the magnitude of the vector.

If you don't know how to use vectors, I'm afraid I won't have much luck helping you understand how to solve this problem. I can't think of an easier way to do it, but I'm a little tired right now.

- Warren
 
Ok that's sound no i don't understand vectors. I'm actually programming it so i mite be able to work another way around it
 
Your model still allows for two possible points. If only two of the points are fixed, then the third can be as you showed in the diagram or it could be reflected across the line described by the two points.

That being said, you could probably figure it out using slopes. Describe the slopes in terms of the points and stipulate that the slopes satisfy the equation m1m2 = -1.

cookiemonster
 
I was thinkin i could work it out by gettibg the area and then workin backwards to the answer but i will still have two unknowns so i don't think that will work properlt
 
Two unknowns? You're going to get two solutions, but you should have only a single unknown.

cookiemonster
 
Well, I really do only see two ways to do this: vector geometry or plain ol' line-equation algebra.

If you have only this one problem to solve, we can give you the solution -- but if you need to write code to solve any triangle like this, you're going to need to bone up on some math. It's pretty tough to write code to do something you don't know how to do by hand.

- Warren
 
  • #10
Two solutions that's what i meant, same as.
 
  • #11
No i need to write code as of now i just have to solution(point) hard coded into it. I want to be able to work this out so as the triangle can be of any size. I vaucly recall doing vector geometry in skul so i'll have a look at that and see if i can recall how it works
 
  • #12
OK i have figured out a way of gettin the x value of the point that i want(see original post), I just got the length of the line between that point and the Hypot. side. I did this by using the hypotneuse as the base and this line as the perpindicular height so now that I have the X value. I think by using the distance formula i should be able to work out Y. Would I be right in assuming this, the only problem is it should return two values for Y the bother is the only Y value I can get is the one that I don't want. Could someone please help, Thanks
 

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