# Thread: Finding the Center of a Right Triangle 3D

1. ## Finding the Center of a Right Triangle 3D

So I'm attempting to find the center of a right triangle in 3d, I've done it before but it's been awhile.

I have three points, p1, p2, p3, all with (x,y,z)
p1 is the top right point. We can give it the points (x,y,z)
p2 will be the bottom right point. We can give it the points (x, y-1, z2)
p3 will be the bottom left point. We can give it the points (x-1, y-1, z3)
The Z coordinates are on the same plane, but can be any integer.
The side lengths are not 1 (because of the angle between z's)
p2 is where the right angle is.

To be honest, If you can solve this for just the (x,y) of the center I'll find the z coordinate. So basically you could just throw the z's out if you want.

Thanks for any help!

2. Originally Posted by ILoveTriangles
So I'm attempting to find the (?) center of a right triangle in 3d, I've done it before but it's been awhile.

I have three points, p1, p2, p3, all with (x,y,z)
p1 is the top right point. We can give it the points (x,y,z)
p2 will be the bottom right point. We can give it the points (x, y-1, z2)
p3 will be the bottom left point. We can give it the points (x-1, y-1, z3)
The Z coordinates are on the same plane, but can be any integer.
The side lengths are not 1 (because of the angle between z's)
p2 is where the right angle is.

To be honest, If you can solve this for just the (x,y) of the center I'll find the z coordinate. So basically you could just throw the z's out if you want.

Thanks for any help!
To answer your question we must know which center you want to calculate:
• the center of the circumcircle
• the center of the incircle
• the centroid of the triangle

3. The coordinates of the centroid of a triangle is just the average of the corresponding coordinates of the vertices.
If the coordinates of the vertces are $(x, y, z_1)$, $(x-1, y,z_2)$, and $(x-1, y- 1,z_3)$ then the coordinates of the centroid are
$\left(\frac{x+ x-1+ x-1}{3}, \frac{y+ y+ y-1}{3}, \frac{z+ z_2+ z_3}{3}\right)=\left(x-\frac{2}{3}, y-\frac{1}{3}, \frac{z_1+ z_2+ z_3}{3}\right)$

4. Originally Posted by ILoveTriangles
So I'm attempting to find the center of a right triangle in 3d, I've done it before but it's been awhile.

I have three points, p1, p2, p3, all with (x,y,z)
p1 is the top right point. We can give it the points (x,y,z)
p2 will be the bottom right point. We can give it the points (x, y-1, z2)
p3 will be the bottom left point. We can give it the points (x-1, y-1, z3)
The Z coordinates are on the same plane, but can be any integer.
The side lengths are not 1 (because of the angle between z's)
p2 is where the right angle is.

To be honest, If you can solve this for just the (x,y) of the center I'll find the z coordinate. So basically you could just throw the z's out if you want.

Thanks for any help!
Are you sure that the point $P_3(x-1, y-1, z_3)$ is written correctly? Shouldn't it be $P_3(x-1, y+1, z_3)$? In the following calculation I'll use these "corrected" values.

The coordinates of the center of the circumcircle in a right triangle (and only there!) are the mean values of the coordinates of the endpoints of the hypotenuse.
According to the text of the question the endpoints of the hypotenuse are $P_2(x, y-1, z_2)$ and $P_3(x-1, y+1, z_3)$

So the center in question has the coordinates

$M\left(\dfrac{2x-1}{2},\dfrac{2y}{2},\dfrac{z_2+z_3}{2} \right)$

This is of course the midpoint of the hypotenuse.