Distance is independent of coordinates I am asked to show $d(x,y) = ((x_2 - x_1)^2 + (y_2 -y_1)^2)^{1/2}$ does not depend on the choice of coordinates. My try is:
$V$ has basis $B = b_1 , b_2$ and $B' = b_1' , b_2'$ and $T = [[a c], [b d]]$ is the coordinate transformation matrix $Tv_{B'} = v_B$ and $x_{B'} = x_1 b'_1 + x_2 b'_2$ and $y_{B'} = y_1b_1' + y_2b_2'$ are the vectors and the distance in the coordinates of $B'$ is $d(x_{B'},y_{B'}) = ((x_2 - x_1)^2 + (y_2 -y_1)^2)^{1/2}$.
The coordinates in $B$ are $x_B = (x_1 a + x_2 c)b_1 + (x_1 b + x_2 d) b_2$ and similar for $y$. I compute the first term in the distance $((x_1 b + x_2 d) - (x_1 a + x_2 c))^2$. I may assume these are Cartesian coordinates so that $a^2 + b^2 = c^2 + d^2 = 1$ and $ac + bd = 0$. 
With this I have $((x_1 b + x_2 d) - (x_1 a + x_2 c))^2 = x_2^2 + x_2^2 - 2(x_1^2 ab + x_1 x_2 bc + x_1 x_2 ad + x_2^2 cd)$. My problem is that $x_1^2 ab + x_1 x_2 bc + x_1 x_2 ad + x_2^2 cd \neq x_1 x_2$. How to solve this? How to show that $x_1^2 ab + x_2^2 cd = 0$ and that $bc + ad = 1$? Thank you.
 A: I would try a little bit more abstract approach. Sometimes a little bit of abstraction helps.
First, distance can be computed in terms of the dot product. So, if you have points with Cartesian coordinates $X,Y$, the distance between them is
$$
d(X,Y) = \sqrt{(X-Y)^t(X-Y)} \ .
$$
Now, if you make an orthogonal change of coordinates of matrix $S$, the new coordinates $X'$ and the old ones $X$ are related through the relation
$$
X = SX'
$$
where $S$ is an orthogonal matrix. This is exactly your condition that the new coordinates are "Cartesian". That is, if
$$
S = \begin{pmatrix}
a   &   c  \\
b   &   d
\end{pmatrix}
$$
the fact that $S$ is orthogonal means that $S^tS = I$, that is
$$
\begin{pmatrix}
a   &   b  \\
c   &   d
\end{pmatrix}
\begin{pmatrix}
a   &   c  \\
b   &   d
\end{pmatrix}
=
\begin{pmatrix}
1   &   0  \\
0   &   1
\end{pmatrix}
\qquad
\Longleftrightarrow 
\qquad
a^2 + b^2 = c^2 + d^2 =1
\quad \text{and} \quad ac + bd = 0 \ .
$$
So, let's now compute:
$$
d(X,Y) = \sqrt{(SX' - SY')^t(SX' - SY')} = \sqrt{(X'-Y')^tS^tS(X'-Y')} = \sqrt{(X'-Y')^t(X'-Y')} \ .
$$
Indeed, distance does not depend on the Cartesian coordinates.
