Given the distance measure for two points and the point $p = (0,2)$, which of the following points have the same distance as $p$ from the origin? Given the distance measure dist:
$$\mathrm{dist}(x,y)  =
\sqrt{(x_1-y_1,x_2-y_2)\begin{pmatrix}
3 & 0\\ 
0 & 4 
\end{pmatrix} (x_1-y_1,x_2-y_2)^{T} } $$
for two dimensional points and the point $p=(0,2)$, which of the following points have the same distance as $p$ from the origin $(0,0)$?

*

*$(4,0)$

*$\color{red}{(2,0)}$

*$(2,1)$

*$(\sqrt{6},0)$

*$(\sqrt{7},0)$

*$\left(\sqrt{\dfrac{16}{3}},0 \right)$

*$\color{red}{(0,-2)}$

The answers in red are my solutions, which I found through this work. However the correct solutions from the Prof. are $(2,1)$ , $\left(\sqrt{\frac{16}{3}}, 0\right)$, and $(0,-2)$.
I just want to know why answers $(2,1)$ and $\left(\sqrt{\frac{16}{3}}, 0\right)$ are correct, it makes no sense to me?
 A: For two points $x = (x_1,x_2)$ and $y =(y_1,y_2)$ we have,
$$\begin{split}
\mathrm{dist}(x,y)  =&
\sqrt{(x_1-y_1,x_2-y_2)\begin{pmatrix}
3 & 0\\ 
0 & 4 
\end{pmatrix} (x_1-y_1,x_2-y_2)^{T} } \\
=& \sqrt{(x_1-y_1,x_2-y_2)\begin{pmatrix}
3 & 0\\ 
0 & 4 
\end{pmatrix}  \begin{pmatrix}
x_1-y_1 \\
x_2-y_2  
\end{pmatrix}  } \\
=& \sqrt{(x_1-y_1,x_2-y_2) \begin{pmatrix}
3(x_1-y_1) \\
4(x_2-y_2)  
\end{pmatrix}  }\\
=& \sqrt{3(x_1-y_1)^2+4(x_2-y_2)^2}
\end{split}$$
Therefore if $y = (0,0)$ then,
$$\mathrm{dist}(x,(0,0)) = \sqrt{3x_1^2+4x_2^2}$$
Hence,
$$\boxed{\mathrm{dist}(p,(0,0)) =\sqrt{3\cdot0^2+4\cdot 2^2}=4}$$
$$\mathrm{dist}((4,0),(0,0)) =\sqrt{3\cdot4^2+4\cdot 0^2}=4\sqrt{3}$$
$$\mathrm{dist}((2,0),(0,0)) =\sqrt{3\cdot2^2+4\cdot 0^2}=2\sqrt{3}$$
$$\boxed{\mathrm{dist}((2,1),(0,0)) =\sqrt{3\cdot2^2+4\cdot 1^2}=4}$$
$$\mathrm{dist}((\sqrt{6},0),(0,0)) =\sqrt{3\cdot\sqrt{6}^2+4\cdot 0^2}=3\sqrt{2}$$
$$\mathrm{dist}((\sqrt{7},0),(0,0)) =\sqrt{3\cdot\sqrt{7}^2+4\cdot 0^2}=\sqrt{21}$$
$$\boxed{\mathrm{dist}\left(\left(\sqrt{\frac{16}{3}},0\right),(0,0)\right) =\sqrt{16+4\cdot 0^2}=4}$$
$$\boxed{\mathrm{dist}((0,-2),(0,0)) =\sqrt{3\cdot0^2+4\cdot (-2)^2}=4}$$
