+0

# Graph

0
67
2

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?

Guest Nov 1, 2017
Sort:

#1
+78643
+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

CPhill  Nov 1, 2017
#2
+78643
+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 )

CPhill  Nov 1, 2017

### 5 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details