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Point of triangle

  1. Apr 5, 2004 #1
    OK I have right angled triangle, I no the lenght of the 3 sides. and the position
    of two of the points but i wanna 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 [Broken]
    Last edited by a moderator: May 1, 2017
  2. jcsd
  3. Apr 5, 2004 #2


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    Do you know how to decompose vectors into components?

    - Warren
  4. Apr 5, 2004 #3
    I dunno wot that is??
    Could u explain???
  5. Apr 5, 2004 #4


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    A vector is a directed line segment -- a distance and a direction.

    The components of a vector [itex]\vec u[/itex] on a plane are [itex](l \cos \theta, l \sin \theta)[/itex] where [itex]l[/itex] 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
  6. Apr 5, 2004 #5
    Ok thats sound no i don't understand vectors. I'm actually programming it so i mite be able to work another way around it
  7. Apr 5, 2004 #6
    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.

  8. Apr 5, 2004 #7
    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
  9. Apr 5, 2004 #8
    Two unknowns? You're going to get two solutions, but you should have only a single unknown.

  10. Apr 5, 2004 #9


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    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 gonna 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
  11. Apr 5, 2004 #10
    Two solutions thats what i meant, same as.
  12. Apr 5, 2004 #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 doin vector geometry in skul so i'll have a look at that and see if i can recall how it works
  13. Apr 6, 2004 #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 lenght 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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