​ HELP ASAP!!

Nevermind, I solved it!

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)

Here's a pic :