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.

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?

*It is not enough to simply plot the points and draw lines -- you must either prove that there is a point that is on all three medians, or show that it is impossible for such a point to exist!*

HelpPLZ Oct 8, 2018

#1**+2 **

A = (4,8) B = (2, -6) C = (-4, 4)

We can use the midpoint theorem to find the medians

median of AB = ( [4 + 2]/ 2, [8 + -6] / 2 ) = ( 3, 1)

median of CA = ( [-4 + 4 ] / 2 , [ 8 + 4 ] / 2 ) = (0, 6)

median of BC = ( [ -4 + 2 ] / 2 , [-6 + 4 ] / 2 ) = (-1, -1 )

Slope of line from A to median of BC = [ 8 - -1] / [ 4 - - 1 ] = 9/5

Equation of the median :

y = (9/5)(x -4) + 8

y = (9/5)x - 36/5 + 40/5

y = (9/5)x + 4/5 (1)

Slope of line from B to median of CA = [ -6 - 6 ] / [ 2 - 0] = -12/ 2 = -6

Equation of the median :

y = -6x + 6 (2)

Slope of line from C to median of AB = [4 - 1 ] / [ -4 -3] = [3 / -7] = [ -3/7 ]

Equation of the median

y = (-3/7)(x - 3) + 1

y = (-3/7)x + 9/7 + 7/7

y = (-3/7)x + 16/7 (3)

Find the x intersection of (1) and (2) by stting these equal

(9/5)x + 4/5 = -6x + 6

(9/5)x + 6x = 6 - 4/5

(9/5)x + (30/5)x = 30/5 - 4/5

39/5x = 26/5

39x = 26

x = 26/39 = 2/3

And plugging this x value into (2) to find y, we have -6(2/3) + 6 = -4 + 6 = 2

So...the intersection point should be ( 2/3 , 2)

Check this in (1) and (3)

(9/5)(2/3) + 4/5 = 18/15 + 12/15 = 30/15 = 2

(-3/7)(2/3) + 16/7 = -6/21 + 48/21 = 42/21 = 2

So.....the medians all pass through ( 2/3 , 2)

CPhill Oct 8, 2018