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In trianlge ABC, AB = 3, AC = 5, and BC = 7. M is the midpoint of BC.  Find the length AM.

 Dec 8, 2019
 #1
avatar+69 
0

HINT

 

I think you can solve this problem by using coordinates. Let BC be the x axis and let B have coordinates of (0,0). So C has coordinates of (7,0). You can find M by doing slopes. M has coordinates of (7/2, 0) because it is the midpoint. So AM can be found by using distance formula.

 Dec 8, 2019
 #2
avatar+106519 
+1

Let A  = (0, 0)   B  =  (3,0)   

 

And we can let C  =  the intersection  of   circles with the equations  of

 

x^2 + y^2  =  25    and   (x -3)^2  + y^2  = 49

 

Subtract the first equation from the second and we have that

 

(x - 3)^2   - x^2   =   24   simplify

 

x^2 - 6x + 9  - x^2  =  24

-6x + 9  = 24

-6x  = 15

x = -15/6  =  -5/2  =  -2.5     this is the x coordinate of C

 

And the y coordinate  =     (-2.5)^2  + y^2  =  25

6.25 + y^2  =  25

y^2   =  25 - 6.25

y^2   = 18.75

y  = sqrt ( 75/ 4)

y = (5/2)√3  = 2.5√3

 

So...C  =    (-2.5, 2.5√3)

 

So....since M is the midpoint of BC....its coordinates are

 

( [ -2.5 + 3)/2 , [ 0 + 2.5√3] / 2 )   =   ( .25, 1.25√3)

 

So

 

AM  =   √[ .25^2  + (1.25√3)^2 ]  = √[ (1/4)^2 + (5√3/4)^2 ]  =  (1/4)√[ 1 + 75]  =  

 

√76  / 4    =   √[ 19 * 4 ] / 4    =   2√19/4    =  √19 / 2

 

Here's a pic :

 

 

cool cool cool

 Dec 8, 2019
edited by CPhill  Dec 8, 2019

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