Question 1147737
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A is point (-1,6) on a cartesian {{{highlight(cross(graph))}}} {{{highlight(plane)}}}, and B is point (14,9) on the same {{{highlight(cross(graph))}}} {{{highlight(plane)}}}. 
Point C is on the x axis. What is the least value of AC+CB?
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<pre>

It follows the motives of well known (famous) minimization problem, solved about 400 years ago by Pierre Fermat.


The solution is as follows.



(1)  Reflect the point A= (-1,6) about the x-axis as if x-axis is a mirror.

     You will get the point A'= (-1,-6).



(2)  Connect the points A'= (-1,-6)  and  the point B= (14,9) by a straight line.

     
     It has the slope m = {{{(9-(-6))/(14-(-1))}}} = {{{(9+6)/(14+1)}}} = 1,

     and its equation is  y - (-6) = 1*(x - (-1)),  which is the same as

     y+6 = x+1,  or  y = x-5.



(3)  Take the x-intercept of this line.

     It is  x= 5, y= 0, i.e. the point  (5,0).

     This point is what the problem asks for :  C = (5,0).


     The distance AC = {{{sqrt((5-(-1))^2 + (0-6)^2)}}} = {{{sqrt(6^2 + 6^2)}}} = {{{6*sqrt(2)}}}.


     The distance BC = {{{sqrt((5-14)^2 + (0-9)^2)}}} = {{{sqrt(9^2 + 9^2)}}} = {{{9*sqrt(2)}}}.


<U>ANSWER</U>.  The point C providing minimum sum of lengths  AC + BC  is  C = (5,0).

         The minimum value of  AC + BC  is  {{{6*sqrt(2)}}} + {{{9*sqrt(2)}}} = {{{15*sqrt(2)}}}.
</pre>

Solved.