What is the center of mass of a triangle with coordinates 0 -12 0 12 1 0?
October 04, 2007 12:14AM
The center of mass of a triangle is better known as the centroid - the intersection of the medians. One median of the triangle with coordinates A(0,-12), B(0,12) and C(1,0) is obvious: the middle of side AB is (0,0) so the line from C to that middle is the x-axis. Another media goes from A to the midpoint of BC. This midpoint M can be expressed as (1/2,6). We want to know the equation of line AM: it's x-intercept, at A, is -12; it goes up 18 units over a horizontal distance of 1/2 a unit, so its slope is 18*2=36. Thus, the equation is y=36x-12. We now solve for the intersection of the two medians: y=0 and y=36x-12. This gives us 36x-12=0 or x=12/36=1/3, so the intersection point is (1/3,0). Theoretically, we could try to find the equation of the third median, but there is no need - we know that it has to pass through the intersection of the first two. So the coordinates of the centroid, or the center of mass, are just this intersection: the point (1/3,0). The coordinates of the center of mass, or centroid, are just the average of the coordinates of the corners. This is a much faster way of calculating the centroid. Our three points are (0,12), (0,-12), and (1,0). The mean of the x-coordinates is (0+0+1)/3 or 1/3. The mean of the y-coordinates is (12-12+0)/3 or 0. So the centroid's coordinates are (1/3,0).