For two given points, we want to find a third point such that the sum of the distances between the original points and the new point is minimized.
We will find that the coordinates of point C are (6, 0).
Remember that the distance between two points K(x₁, y₁) and N(x₂, y₂) is given by:
[tex]KN = \sqrt{(x_2 - x_1)^2 +(y_2 - y_1)^2}[/tex]
Now for the given points:
A(3, 6)
B(8, 4)
We want to find a point C on the x-axis such that:
AC + BC is minimized.
Because C is on the x-axis, the y-value is zero, then we can write:
C(x, 0).
Using the equation for the distance, we can write:
[tex]D = AC + BC = \sqrt{(3 - x)^2 + 6^2} + \sqrt{(8 - x)^2 + 4^2} \\\\D = \sqrt{(3 - x)^2 + 36} + \sqrt{(8 - x)^2 + 16}\\\\D = \sqrt{9 - 6x + x^2 + 36} + \sqrt{64 -16x + x^2 + 16}\\\\D = \sqrt{x^2 - 6x + 45} + \sqrt{x^2 - 16x + 80}[/tex]
So we want to minimize D, to do it, we need to see the zeros of the derivate of D.
The derivate of D is given by:
[tex]D = (x^2 - 6x + 45)^{1/2} + (x^2 - 16x + 80)^{1/2}\\\\D' = (1/2)*(x^2 - 6x + 45)^{-1/2}*(2x - 6) + (1/2)*(x^2 - 16x + 80)^{-1/2}*(2x - 16)\\\\D' = (x^2 - 6x + 45)^{-1/2}*(x - 3) + (x^2 - 16x + 80)^{-1/2}*(x - 8)\\\\D' = \frac{x-3}{\sqrt{(x^2 - 6x + 45)} } + \frac{x -8}{\sqrt{ (x^2 - 16x + 80)} }[/tex]
So we need to find the zeros of D'.
Doing this analitically is possible, but the calculations would be too long (we would need to solve a quartic equation). Then we can solve it graphically.
Below you can see the graph of the equation of D'
There, you can see that it crosses the x-axis at x = 6.
Thus the value of X that minimizes D is x = 6
Then the coordinates of point C are: (6, 0)
If you want to learn more, you can read:
https://brainly.com/question/23397675
