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
Computer Science Canada cetroid finder |
| Author: | Jimbo 420 [ Thu Jun 08, 2006 11:46 am ] |
| Post subject: | cetroid finder |
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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| Author: | upthescale [ Thu Jun 08, 2006 7:37 pm ] |
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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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| Author: | Delos [ Thu Jun 08, 2006 9:11 pm ] |
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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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| Author: | upthescale [ Thu Jun 08, 2006 10:00 pm ] |
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how is that sterotype delos? |
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| Author: | Clayton [ Thu Jun 08, 2006 10:09 pm ] |
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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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| Author: | Jimbo 420 [ Mon Jun 12, 2006 11:48 am ] |
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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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| Author: | TheOneTrueGod [ Mon Jun 12, 2006 1:07 pm ] |
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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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| Author: | Jimbo 420 [ Wed Jun 14, 2006 11:59 am ] |
| Post subject: | Nice |
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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| Author: | Jimbo 420 [ Wed Jun 14, 2006 12:06 pm ] |
| Post subject: | more help |
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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