Question

Consider triangle ABC below note that triangle ABC has verticals A(5,-7) B(-3,5) and C(-7,-3) Suppose a D is the midpoint of AB and the E is a midpoint of AC answer the following

Answer 1

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 )

Explanation:

Got u fam and i have done this before