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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?

 Dec 5, 2017
edited by Guest  Dec 5, 2017
 #1
avatar+7350 
+1

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.

 Dec 5, 2017
 #2
avatar+98044 
+3

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  ]

 

 

cool cool cool

 Dec 5, 2017
 #3
avatar+7350 
+1

Ohhh I thought there might be a better way to do it...   smiley

hectictar  Dec 5, 2017

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