A median of a triangle is a line segment joining a vertex of a triangle to the midpoint of the opposite side. The three medians of a triangle are drawn below. Note that the three medians appear to intersect at the same point! Let's try this out with a particular triangle. Consider the triangle ABC with A = (3,6), B = (-5,2), and C = (7,-8). (a) Let D, E, F be the midpoints of \overline{BC}, \overline{AC}, \overline{AB}, respectively. Find the equations of medians \overline{AD}, \overline{BE}, and \overline{CF}. (b) Show that the three medians in part (a) all pass through the same point.
https://latex.artofproblemsolving.com/b/c/3/bc36d6a86ebdd87399c8b07d66ec053e8593c264.png
+128475 Hard to see this....but
Midpoint of AB = [ (3 -5)/2, (6 + 2) / 2 ] = [ -1, 4 ] = F
Midpoint of of BC = [ (-5 + 7)/2 , (2- 8)/2 ] = [ 1 , -3] = D
Midpoint of AC = [ (3 + 7)/2 , ( 6 - 8)/2 ] = [ 5, -1] = E
Slope of CF = [ -8 - 4 ] / [ 7 - -1] = -12/8 = -3/2
Equation of line containing CF =
y= -(3/2) ( x - - 1) + 4
y = -(3/2) ( x + 1) + 4
y= (3/2)x - 3/2 + 4
y = -(3/2)x + 5/2 (1)
Slope of AD = [ 6 - -3 ] / [ 3 - 1] = 9/ 2
Equation of line containing AD =
y = (9/2) (x -3) + 6
y =(9/2)x - 27/2 + 12/2
y= (9/2)x - 15/2 (2)
We can the find x intersection of medians CF and AD by setting (1) = (2)....so we have
-(3/2)x + 5/2 = (9/2)x - 15/2
10 = 6x
5 = 3x
x = 5/3
And using (1) the y value of the intersections is
y = -(3/2)(5/3) + 5/2 = 0
So....the intersection of the medians = (5/3 , 0)
We can check this by writing an equation for the remaining median, BE
Slope of line containing BE =
[ 2 - -1 ] / [ -5 - 5] = 3 / -10 = -3/10
So the equation of this line is
y = -(3/10)(x - - 5) + 2
y = - (3/10)x - 15/10 + 20/10
y= -(3/10)x + 5/10
y = -(3/10)x + 1/2
Note that when x = (5/3) we have that
y = -(3/10)(5/3) + 1/2
y = - (5/10) + 1/2
y= -1/2 + 1/2
y = 0
So.....this confirms that the intersction of the medians = ( 5/3, 0 )
