Answer
See explanations.
Work Step by Step
It is expected that the distance between two points is invariant under rotation because the distance should only depend on the relative locations of the two points, not the axes of the coordinates. To prove this, you can do the following steps:
Step 1. Assume the coordinates of points P and Q as $P(x,y)$ and $Q(u,v)$ in the xy-system, and their coordinates change to $P(X, Y)$ and $Q(U, V)$ in the XY-system after rotation.
Step 2. Write the relationship between the above coordinates:
$x=Xcos\phi-Ysin\phi$, $y=Xsin\phi+Ycos\phi$. $u=Ucos\phi-Vsin\phi$, $v=Usin\phi+Vcos\phi$
Step 3. The distance in the xy-system is given by $d=\sqrt {(x-u)^2+(y-v)^2}$ and the distance in the XY-system is given by $D=\sqrt {(X-U)^2+(Y-V)^2}$
Step 4. Use the relationships in step-2 in the $d$ formula:
$d^2=(Xcos\phi-Ysin\phi-Ucos\phi+Vsin\phi)^2+(Xsin\phi+Ycos\phi-Usin\phi-Vcos\phi)^2=[(X-U)cos\phi-(Y-V)sin\phi]^2+[(X-U)sin\phi+(Y-V)cos\phi]^2$
Step 5. Use the Identify $sin^2\phi + cos^2\phi =1$, expand the above expression, we get $d^2=D^2$ and $d=D$, thus we prove that the distance is invariant under axes rotations.
