2021/099) Find the minimum value of $\sqrt{​x^2+​4​x+​13}+​\sqrt{​x^2-​8​x+​41}$

 We have 

$\sqrt{​x^2+​4​x+​13}+​\sqrt{​x^2-​8​x+​41}$

= $\sqrt{​(x+​2)^2+​9}+​\sqrt{​(x-4)^2 + 25}$

The 1st term that is $\sqrt{​(x+​2)^2+​9}$ is distance from (x,0) to (-2,-3) and second term $\sqrt{​(x-4)^2+25}$ is distance from (x,0) to (4,-5).

clearly sum of the distanace to (x,0) is lowest when (x,0), (-2,3) and (4,5) are in a straight line that is (x,0) is on the line from (-2,3) to (4,5)  now as (x,0) lies in between the minimum is that distance from (-2.-3) to (4,5) or $\sqrt{(4+2)^2 + (5+3)^2} = \sqrt{100} = 10$