SOLUTION: 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. What is the least value of AC+CB?

Algebra ->  Coordinate-system -> SOLUTION: 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. What is the least value of AC+CB?      Log On


   



Question 1147737: 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. What is the least value of AC+CB?
Answer by ikleyn(52787) About Me  (Show Source):
You can put this solution on YOUR website!
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A is point (-1,6) on a cartesian highlight%28cross%28graph%29%29 highlight%28plane%29, and B is point (14,9) on the same highlight%28cross%28graph%29%29 highlight%28plane%29.
Point C is on the x axis. What is the least value of AC+CB?
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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 = %289-%28-6%29%29%2F%2814-%28-1%29%29 = %289%2B6%29%2F%2814%2B1%29 = 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%28%285-%28-1%29%29%5E2+%2B+%280-6%29%5E2%29 = sqrt%286%5E2+%2B+6%5E2%29 = 6%2Asqrt%282%29.


     The distance BC = sqrt%28%285-14%29%5E2+%2B+%280-9%29%5E2%29 = sqrt%289%5E2+%2B+9%5E2%29 = 9%2Asqrt%282%29.


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

         The minimum value of  AC + BC  is  6%2Asqrt%282%29 + 9%2Asqrt%282%29 = 15%2Asqrt%282%29.

Solved.