A B C D E F 500ft <-> 500ft <-> 500ft <-> 500ft <-> 500ft <-> 2000ft
Let's say the bus stops at A = 0 ( from A ) + 500 ( from B ) + 500 + 500 ( from C ) + 500 + 500 + 500 ( from D ) + 500 + 500 + 500 + 500 ( from E ) + 500 + 500 + 500 + 500 + 2000 ( from F ) = 0 + 500 + 1000 + 1500 + 2000 + 4000 = 9000 ft
at B = 500 ( from A ) + 0 ( from B ) + 500 ( from C ) + 500 + 500 ( from D ) + 500 + 500 + 500 ( from E ) + 500 + 500 + 500 + 2000 ( from F ) = 500 + 0 + 500 + 1000 + 1500 + 3500 = 7000 ft
at C = 500 ( from A ) + 500 + 500 ( from B ) + 0 ( from C ) + 500 ( from D ) + 500 + 500 ( from E ) + 500 + 500 + 2000 ( from F ) = 500 + 1000 + 0 + 500 + 1000 + 3000 = 6000 ft
Do the same for D , E and F
at D = 6000 ft
at E = 7000 ft
at F = 15000 ft
You will find that the bus should stop at either C or D to make the sum of distances from every house to the stop as small as possible.
Please let me know if you want me to add an explanation as to why this is the answer. I can definitely do that, I just wouldn’t want to write it if you don’t want me to :)