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A is point (-1,6) on a cartesian , and B is point (14,9) on the same .
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 = = = 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 = = = .
The distance BC = = = .
ANSWER. The point C providing minimum sum of lengths AC + BC is C = (5,0).
The minimum value of AC + BC is + = .
Solved.