CPhill

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 #3
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+2
1 jul. 2020
 #3
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+2

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 :

 

 

 

cool cool cool

1 jul. 2020
 #14
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+3
22 may. 2020
 #1
avatar+111437 
+2

We  can find  the height  of  this triangle  as  follows

 

Using  Heron's Formula  we have  the area as

sqrt  (9 (9-8) (9-6) (9 - 4))  = sqrt  ( 9 * 3 * 5)  =  3sqrt(15)

So.....to find the height we  have

Area  = (1/2)(BC) height

3sqrt (15)  =  (1/2)(4) height

3sqrt (15)  = 2 * height

(3/2)sqrt (15)  = 1.5 sqrt (15)  =  height  = y coordinate of A

 

And we  can find  the  x coordinate  of A  by the Pythagorean Theorem

 

sqrt  [ AC^2  -  height of ABC^2 ]   = sqrt  (8^2   -(1.5 sqrt (15))^2  )  = sqrt (30.25)  =  5.5

 

So....the coordinates  of A  = (5.5, 1.5sqrt (15))

 

 

We  can use a formula to find  the coordinates  of  the center  of the incircle

Let B = (4,0)  and C  = (4,0)

 

x coordinate  of incenter  =

 

[ Ax* a  + Bx* b  + Cx * c ]  / perimeter

 

Where Ax = the x coordinate  of A  Bx  = x coordinate of B    Cx  = x coordinate of C

And a, b , c  are  the sides lengths opposite A, B and C

 

So  we have

 

[ 5.5 (4)  + (4)(8)  + (0)(6)] / 18  =  3

 

Similarly the  y coordinate  of the center of  the incircle  =  

[Ay * a  + By * b + Cy * c ]  / 18 

 

[ 1.5sqrt (15)(4)  + (0)(8) + (0)(6)  / 18   =  sqrt (15)/3   =    radius  of incircle

 

Then the  height of triangle ANM = height of triangle ABC  - 2* incircle radius = sqrt (15)  ( 3/2 - 2/3)  =

 

(5/6) sqrt (15 )

 

And since triangles  AMN and ABC  are similar

 

Then

 

MN  / height of AMN  =  BC / height of ABC

 

MN  / [ (5/6)sqrt (15) ] =   4 /  [  (3/2 ) sqrt (15) ]

 

MN =  4 (5/6)  / (3/2)   =  4 ( 5/6) (2/3)  =  40 /18  =  20/9  

 

 

cool cool cool

22 may. 2020