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a is point (-1,6) on a cartesian graph and b is point (14,9) on the same graph. Point C is on the x axis. The least value of line ac plus line cb is?

 Nov 1, 2017
 #1
avatar+98197 
+1

Let the point  c on the x axis be  (x, 0)

 

ac + cb  can be represented as D =

 

D  =  sqrt [ (x + 1)^2 + 6^2 ]  + sqrt [ (14- x)^2 + 9^2]

 

D = [  x^2 + 2x  + 37 ]^(1/2) + [ x^2  - 28x + 277 ]^(1/2)

 

Take the derivative and set to 0

 

D'  =  [ 2x + 2] / ( 2  [  x^2 + 2x  + 37 ]^(1/2))  +  [ 2x - 28] / (2   [ x^2  - 28x + 277 ]^(1/2) )  = 0

 

[ x + 1 ] /   [  x^2 + 2x  + 37 ]^(1/2)  +  [ x - 14] / [ x^2  - 28x + 277 ]^(1/2)  = 0

 

[ x + 1 ] /   [  x^2 + 2x  + 37 ]^(1/2)  =   [ 14 - x ] /  [ x^2  - 28x + 277 ]^(1/2)

 

Square both sides

 

[ x^2 + 2x + 1 ] /  [  x^2 + 2x  + 37 ] =  [ x^2 - 28x + 196] /  [  x^2 -28x + 277 ]

 

Cross - multiply

 

[ x^2 + 2x + 1 ] [  x^2 -28x + 277 ]  =  [ x^2 - 28x + 196]  [  x^2 + 2x  + 37 ]

 

Simplify

 

x^4 - 26 x^3 + 222 x^2 + 526 x + 277  =  x^4 - 26 x^3 + 177 x^2 - 644 x + 7252

 

45x^2  + 1170x  - 6975  =  0

 

9x^2  + 234x -  1395  =  0 

 

Solving this for  x produces   x = 5  or x = -31  [ reject the second solution ]

 

So.....the distance is minimized when   c  =   (5, 0 )

 

We can prove this if

 

arctan  [  6/[5- -1] ]  =  arctan [ 9 / [ 14 - 5] ]   is true

 

arctan  [  6/6]  =  arctan [ 9/9]

 

arctan 1  =   arctan 1         and it is true

 

And the minimum distance  ac + cb  is

 

sqrt [ (5 + 1)^2 + 6^2 ]  + sqrt [ (14- 5)^2 + 9^2] =

 

√ 72  +  √ 162  =

 

√ 2  [ 6 + 9]  =

 

15√ 2  units  

 

 

 

 

 

cool cool cool

 Nov 1, 2017
 #2
avatar+98197 
+1

Alternatively.....if you haven't had Calculus....you could solve this directly for x

 

6 / [ x + 1 ] =  9 / [ 14 - x ] 

 

Cross-multiply

 

6 [ 14 - x ]  = 9 [ x + 1 ]

 

84 - 6x  =   9x + 9

 

75  = 15x

 

 x  = 5       =  c  =   (5, 0 )    

 

 

cool cool cool   

 Nov 1, 2017

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