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Jimbo 420
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Posted: Thu Jun 08, 2006 11:46 am Post subject: cetroid finder |
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this program is suposed to find the centroid of a triangle i can get 2 line eqasions but can't get it to find the centroid need help
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centroid finder.t |
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1.6 KB |
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128 Time(s) |
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upthescale
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Posted: Thu Jun 08, 2006 7:37 pm Post subject: (No subject) |
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are you asking for help? go the the main menu, and go in Turing Help, and post this same post there, because this is users submisions, not help quesitons!
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Delos
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Posted: Thu Jun 08, 2006 9:11 pm Post subject: (No subject) |
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Moved.
And Jimbo 420, please change your sig. Statements promoting negative stereotypes, racism, or any other form of discriminatory thought are not tolerated. Even in a humorous form. I'm sure you did not mean any harm by it, but it still is really pushing the envelope.
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upthescale
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Posted: Thu Jun 08, 2006 10:00 pm Post subject: (No subject) |
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how is that sterotype delos?
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Clayton
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Posted: Thu Jun 08, 2006 10:09 pm Post subject: (No subject) |
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^the above post is spam^ you have been warned too many times to count and it is really annoying plz stop, it is a stereotype because it is using Jews (joo) as a point of humour (and improperly i might add)
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Jimbo 420
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Posted: Mon Jun 12, 2006 11:48 am Post subject: (No subject) |
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any1 know how to finish this program it gives the 2 lines and the 2 lines must be solved to find the answer
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TheOneTrueGod
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Posted: Mon Jun 12, 2006 1:07 pm Post subject: (No subject) |
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Theres two methods.
A) Substitution
L1: 2X + 4Y = 6
L2: 3X - 6Y = 12
from L1: X = (6-4Y)/2
X = 3 - 2Y
sub into L2:
3(3-2Y) - 6Y = 12
9 - 6Y - 6Y = 12
-6Y = 3
Y = -1/2
Sub back into the L1 eqn
2X + 4(-1/2) = 6
2X - 2 = 6
2X = 8
X = 4
so the point of intersection is (4,-1/2)
B)Add the equations
L1: X + 3Y = 5
L2: X - 4Y = 3
L1 - L2 : X-X + 3Y - (-4Y) = 5 - 3
3Y + 4Y = 2
7Y = 2
Y = 2/7
Now continue by subbing it in to the equations.
Summary
You'd probably be better off using the first method, though you'd have to be specific as to the input.
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Jimbo 420
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Posted: Wed Jun 14, 2006 11:59 am Post subject: Nice |
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yeh i had something like that going on thanks for the help tho im pretty sure it works good now thanks man
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Jimbo 420
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Posted: Wed Jun 14, 2006 12:06 pm Post subject: more help |
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ok i can't seem to figure out how to code the substitution equasions? would it work easier using a procedure(which i havent used many of:|)
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