I need help really fast I don't know how to even start!
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+128475 A = (3,6) B = (-5,2) C = (7, -8)
Midpoint of BC = [ (7 - 5) / 2 , (-8 + 2) / 2) = ( 2/2 , -6/2) = (1, -3) = D
Slope of line through AD = [ 6- - 3 ] / [ 3 - 1 ] = 9/2
Equation of line through AD ... y = (9/2) ( x -3) + 6 ......y = (9/2) x - 27/2 + 6 ......y = (9/2)x - 15/2
Midpoint of AC = [ (3 + 7)/2 , (-8 + 6))/2 ] = (10/2, -2/2) = ( 5, -1) = E
Slope of line through BE = [ -1 - 2 ] / [ 5 - -5] = -3/10
Equation of line through BE....y= (-3/10) ( x-5) - 1 ......y = (-3/10)x + 15/10 - 1 .....y = (-3/10)x + 1/2
Now we can find the x intersection of these mediians by setting the equations of these lines equal
(9/2) x - 15/2 = (-3/10)x + 1/2
(9/2 + 3/10) x = (1/2 + 15/2)
(24/5)x = 8
x = 8 (5/24) = 40/24 = 5/3
And the y coordinate of the intersection of these medians is
y = (-3/10)(5/3) + 1/2 = 0
So.....the intersection of these medians = (5/3 , 0)= G
Now....the midpoint of AB = [(-5+ 3)/2, (2+6)/2) ] = (-2/2, 8/2) = (-1,4) = F
And ths slope thruogh FC = [ -8- 4]/ [7- - 1] = [ -12/8] = -3/2
So the equation of the line through FC .....y = (-3/2) (x - -1) + 4.....y = (-3/2)x - 3/2 + 4....y = (-3/2)x +5/2
So...to prove that G is on this line...let x = (5/3)
So when x =(5/3) then y = (-3/2)(5/3) + 5/2 = -5/2 + 5/2 = 0
So ( 5/3 , 0) is on FC whivh proves that all therr medians intersect at G = (5/3, 0)
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