Question

A(2,-2) B(11,-4), find point c on x-axis so AC+BC is a minimum

Answer 1

Call x the x-coordinate of C.
Distance: (AC)^2 = (x-2)^2/4 = (x^2 - 4x + 4)/4
Distance: (CB)^2 = (11 - x)^2/16 = (121 - 22x +x^2)/16
AC + CB minimum when (AC)^2 + (CB)^2 minimum
(AC)^2 + (CB)^2 = (4x^2 - 16x + 16)/16 +(121 - 22x + x^2)/16
= (5x^2 - 38x + 237)/16
The function f(x) = 5x^2 - 38x + 237 is an upward parabola. Its minimum occurs when x = -b/2a (parabola axis)
x = 38/10 = 3.8