# Find the set of points in euclidian and taxicab metric that are the same distance from $(0,1)$ and $(1,0)$

Find the set of points in euclidian and taxicab metric that are the same distance from A = $$(0,1)$$ and B =$$(1,0)$$.

So I just started doing metric spaces and I have got to this simple problem.

My understanding:

We can define a metric as:

$$d((x_1,x_2,...,x_n),(y_1,y_2,...,y_n)) = [\sum_{i=1}^{p}|x_i-y_i|^{1/p}]^p$$ , where if $$p = 1$$ we call it taxicab metric and if $$p = 2$$ we call it euclidian metric.

However for this problem I do not know how to begin it. Do I have to look the taxicab metric as only in $$\mathbb{R^1}$$ space and thus we have a point $$x_1 = 1$$ and a point $$y_1 = 0$$ and then for euclidian metric as in $$\mathbb{R^2}$$ and $$x_1 = 1, x_2 = 0$$ and $$y_1 = 0, y_2 = 1$$ .

My question is how to go about problems like these?

• Start with the obvious solutions (0,0) and (1,1) and show there aren't any others. Commented Aug 6, 2021 at 16:38
• @herbsteinberg: There are plenty others. For example $(t,t)$ works under both metrics for every $t\in\mathbb R$. Commented Aug 6, 2021 at 18:39
• OP (@VLC), I think you might be led astray by a typo in the definition -- the summation should be $\sum_{i=1}^n$ rather than $\sum_{i=1}^p$. The dimension is independent of with $p$ you use, so both Euclidean and taxicab/Manhattan metric can work for $\mathbb R^2$. Taxicab is then just $$d_1((x_1,x_2),(y_1,y_2)) = |x_1-y_1|+|x_2-y_2|$$ or -- with perhaps more intuitive variable names -- $$d_1((x_A,y_A),(x_B,y_B)) = |x_A-x_B|+|y_A-y_B|$$ Commented Aug 6, 2021 at 18:41
• @Troposphere The original question is somewhat ambiguous. My interpretation is that both metrics would give the same distances, i.e. $d_t(p.A)=d_e(p.A)$ and $d_t(p.B)=d_e(p.B)$ Commented Aug 6, 2021 at 20:38
• @herbsteinberg: Not quite. For exampe if $P=(5,3)$ then $d_t(P,A)=d_t(P,B)=7$ but $d_e(P,A)=3\sqrt 3$ and $d_e(P,B)=5$. Commented Aug 6, 2021 at 21:28

These are the points $$(x,y)$$ such that $$d((x,y),(0,1))=d((x,y),(1,0).$$ That is, $$|x|+|y-1|=|x-1|+|y|$$