A median of a triangle is a line segment from a vertex of a triangle to the midpoint of the opposite side of the triangle. Below, the three medians of the black triangle are shown in red.
https://latex.artofproblemsolving.com/1/5/5/15593c90ce20b22893a1bd76ac6d5f6e3b61f5bb.png
Notice that the three medians appear to pass through the same point! Let's test this out with a specific triangle. Consider one specific triangle $ABC$ with $A = (4,8)$, $B = (2,-6)$, and $C = (-4,4)$. Do the three medians of this triangle pass through the same point? If so, what point?
+9466 the midpoint of AB = \((\frac{4+2}{2},\frac{8-6}{2})\) = (3, 1)
the midpoint of AC = \((\frac{4-4}{2},\frac{8+4}{2})\) = (0, 6)
the midpoint of BC = \((\frac{2-4}{2},\frac{-6+4}{2})\) = (-1, -1)
a line that passes through (3, 1) and (-4, 4) has the equation \(y=-\frac37x+\frac{16}{7}\)
a line that passes through (0, 6) and (2, -6) has the equation \(y=-6x+6\)
a line that passes through (-1, -1) and (4, 8) has the equation \(y=\frac95x+\frac45\)
The first two lines intersect when
\(-\frac37x+\frac{16}{7}=-6x+6 \\ -\frac37x+6x=6-\frac{16}{7}\\ \frac{39}{7}x=\frac{26}{7} \\x=\frac{2}{3}\)
And \(y=-6(\frac23)+6=2\)
The first two lines intersect at the point \((\frac23,2)\) . So if \((\frac23,2)\) is on the third line, then we know that all three lines pass through \((\frac23,2)\) .
\(y=\frac95(\frac23)+\frac45=\frac65+\frac45=\frac{10}{5}=2\)
So...all three medians pass through \((\frac23,2)\) . Here's a graph of the three lines.
+128448 Thanks, hectictar....
The intersection point of the three medians is called the centroid
There's an easy way to figure this......
Median = ( sum of the x coordinates of the vertices / 3 , sum of y coordinates of the vertices / 3 ) =
[ (4 + 2 - 4 ) / 3 , ( 8 - 6 + 4 ) / 3 ] =
[ 2 /3 , 2 ]
