Jenrak, once again, is an idiot at math and requires aid.
I have to finish a summative assessment tommorrow, and I would just like to confirm, that to find a centroid of a shape, do I have to flip the reciprocal and such?
For example: If A(2,3) and B (4,6) C (4,1) (I'm making this up)
Mp = 2+4, 3+4
.........2 ..... 2
= 3, 3.5
And then for slope:
6-3
4-2
=3
...2
Should I flip the slope and change the sign so it's -2 ?
3
Kryozerkia
15-06-2005, 23:57
Sure what the hell...
German Nightmare
16-06-2005, 00:29
Okay. First of all: Is the "centroid" of a shape (in this case a triangle) its "weight-centerpoint"?
You'd have to use all three X/Y coordinates divided by three, I believe?
A(2/3) B(4/6) C(4/1)
----(2+4+4)--(3+6+1)
Z=(-------- / --------)
-------3---------3
Z=(3.333/3.333)
That's what I came up with by drawing it as well.
Then what do you mean by slope?
The slope of what?
O(0/0) to Middlepoint Z(3.333/3.333)?
NAh, it's jsut whether you flip the slope and make it a reciprocal, cus no reliable source tells me to do anything.
Celtlund
16-06-2005, 02:42
Vinie Vertians oxyetarianized the fobanizer in the gunia gista down by the old gas house.
Blessed Misfortune
16-06-2005, 02:44
I suck at math, too. :(
Chocolate is Yummier
16-06-2005, 02:52
I have to finish a summative assessment tommorrow, and I would just like to confirm, that to find a centroid of a shape, do I have to flip the reciprocal and such?
For example: If A(2,3) and B (4,6) C (4,1) (I'm making this up)
Mp = 2+4, 3+4
.........2 ..... 2
= 3, 3.5
And then for slope:
6-3
4-2
=3
...2
Should I flip the slope and change the sign so it's -2 ?
3
i can't exactly remember (i did this a few weeks ago and i've already forgotton, tsk, tsk) but if slope is rise/run then wouldn't flipping it over like that make it a negative slope? I think what you've is okay but don't take my word for it, it might be completely wrong for all i know.
Well I'm trying to find the median, so anyone can tell me?
If not, then I shalt follow Chocolate is Yummier's advice and go that way....
Robot ninja pirates
16-06-2005, 02:56
I have to finish a summative assessment tommorrow, and I would just like to confirm, that to find a centroid of a shape, do I have to flip the reciprocal and such?
For example: If A(2,3) and B (4,6) C (4,1) (I'm making this up)
Mp = 2+4, 3+4
.........2 ..... 2
= 3, 3.5
And then for slope:
6-3
4-2
=3
...2
Should I flip the slope and change the sign so it's -2 ?
3
I doubt it, since the negative reciprocal of a slope is the slope of a line perpindicular to it.
I don't know this stuff, but if you're trying to find the center of 3 points maybe take the average of the x coordinates and the y coordinates.
-edit- for points A(x1,y1) and B(x2,y2) the slope is (y2-y1)/(x2-x1) and the midpoint is M((x1+x2)/2, (y1+y2)/2)
Chocolate is Yummier
16-06-2005, 02:57
this is so deppressing, we spent ages on this and i can't remember anything! i'll keep thinking though.
Leonstein
16-06-2005, 03:39
Bugger. I used to be good at math in school, and all i use now is bloody statistics and matrices. And some algebra.
But never geometry.
Pschycotic Pschycos
16-06-2005, 03:52
Wow...and I thought 10th grade geometry was hard...
lol. I need knowledge....
Yes. Yes, you want to take the negative reciprocal. Next, you use that to create a new line equation, that of the perpendicular bisector of AB. Which would be...
y - 3.5 = -2/3 (x - 3)
Next, do it again with BC (or AC), make another equation, equal the one you make with the one I just gave you, solve and voila.
Say if you don't understand. I rarely make a whole lot of sense.
Grand Teton
16-06-2005, 21:16
Yes. Yes, you want to take the negative reciprocal. Next, you use that to create a new line equation, that of the perpendicular bisector of AB. Which would be...
y - 3.5 = -2/3 (x - 3)
Next, do it again with BC (or AC), make another equation, equal the one you make with the one I just gave you, solve and voila.
Say if you don't understand. I rarely make a whole lot of sense.
Yes, that is logical, those two lines point of intersection would be at the centrepoint of the triangle. Congrats, you are officially a maths teacher!
Yes, that is logical, those two lines point of intersection would be at the centrepoint of the triangle. Congrats, you are officially a maths teacher!
w00t!
I'll be expecting payment... :p
Yeah Okay thanks.
No problem. Some of us are sad and like Maths. :cool:
What class is this for?
(if you tell me "math" I'm gonna kick your ass - what level?)
No problem. Some of us are sad and like Maths. :cool:
:cool:
It occurred to me after I went offline last night: I can't really remember the definition of a centroid. It's either the point of intersection of two perpendicular bisectors or it's the point of intersection between two medians.
So if you're wrong, it's not my fault. :)