Answer:
Shortest distance=|AC|/[tex]\sqrt[2]{2}[/tex].
Kindly find the attached for the figure
Step-by-step explanation:
This problem can be addressed using right-angled,
Let have right angle triangle with the shortest distance=hypotenuse;
hypotenuse=|AC|
using pythagoras theorem, we have
|AC|^2=|AB|^2+|BC|^2
Let |AB|=|BC|
|AC|^2=2|AB|^2
|AB|=|AC|/[tex]\sqrt[2]{2}[/tex]
Shortest distance=|AC|/[tex]\sqrt[2]{2}[/tex]
