The Homework Thread

Downed, I know there's really no way in words to describe it, (off the top of my head atleast)
An isosceles triangle measures out to split into 3 lengthwise. the point between the 1/3 and the 2/3 is called the centroid. He is supposed to find the centroid as his answer.

I decided this could be used to find one of the few ways to find the centroid not in graph or picture form.
 
You forgot to all-caps "she". :p

There's a reason why people don't ask her for help a lot. Almost every few seconds she makes a mistake that messes up the problem, and we have to redo it.
 
amaterasu said:
I have revived this thread from almost a month of death. I hope nothing happens to me.

So, I recently came upon a really confusing topic in Geometry (or it seemed to be confusing). But our teacher taught it so bad. So I'm trusting the people at T9K for help (make sure I don't have any regrets). I'm putting two questions out of the lesson packet she gave us onto here, so I can get a good idea of how to do this stuff.

Find the coordinates of the centroid of the triangle with the given vertices.

X(-3, 15), Y(1,5), Z(5,10)

Find the coordinates of the orthocenter of the triangle with the given vertices.

L(8,0), M(10, 8), N(14, 0)

Hope you can help.
Click to expand...
Doing this now, because I actually know what you want. Will edit shortly.

1. I googled "centroid" because I don't remember which lines that wants. Apparently it wants the "midpoint to opposite vertex" lines. So, let's find the equations of two of these lines.
1a. First we need a midpoint. ((x1+x2/2) , (y1+y2/2)) = ((-3+1)/2 , (15+5)/2) = (-1,10)
1b. Find the line including this sparkly new midpoint, and the unused vertex. This will be one of our centroid lines.
-1,10 5,10
y=mx+b
m= ... 0. That was anticlimatic.
Great, the line is x=10. Let's find another line.
How about 1,5 and 5,10. Midpoint time.
1a. (3,15/2)
1b.

Why am I doing your homework for you? Here is how you do it. (I think. It has been a while since geometry)
Find midpoint of a line.
Use that midpoint and the remaining vertex to find the equation of a new line. This line will pass through the centroid.
Repeat the last two steps, using either of the two remaining triangle lines (not your new one).
Now you have two new line equations, both of which pass through the centroid. Find where they are equal.
(Solve for a variable, substitute, solve for remaining variable, plug that back into original, you know the drill.)
You now have the cords for the centroid. Congrats.

Second problem in progress...

Oh boy, this looks fun. Maybe a bit easier, though.
Find the slope of a triangle line.
Invert said slope. (m= 2 becomes m= -1/2. Not sure if 'invert' is the correct word. Find the perpendicular slope.)
Find the equation of a line with that slope, containing the third as-of-yet-unused-vertex.
Repeat above steps.
You now have two new lines. They are perpendicular to a triangle line and pass through its opposite vertex.
Find where lines are equal. Same as the last problem.
 
konflakes said:
1 + 1 = 3
Click to expand...

Jackie-chan-meme.png
 
You must log in or register to reply here.
Back
Top