Assume one such line exists. First, scale this line so that the the coordinates are integers. Now, we can represent each edge of this broken line as a vector $ (x,y)$ with $ x$ and $ y$ and magnitude $ d$. The sum of all these vectors will be $ (0,0)$. Call these vectors $ (x_1,y_1),(x_2,y_2),...,(x_n,y_n)$. where $ n$ is odd. Then $ x_1^2 + y_1^2 = x_2^2 + y_2^2 = ... = x_n^2 + y_n^2 = d^2$. Suppose that at least one of $ x_1,x_2,...,x_n,y_1,y_2,...,y_n$ is odd. Otherwise, we can divide all of the coordinates and $ d$ by $ 2$. Now, because $ x_1 + x_2 + \cdots + x_n = 0$ and $ y_1 + y_2 + \cdots + y_n = 0$, an even number of $ x_1,x_2,...,x_n$ are odd and an even number of $ y_1,y_2,...,y_n$ are odd. Now I will show that $ d$ is even. Suppose for the sake of contradiction that $ d$ is odd. Then for each pair $ (x_i,y_i)$, exactly one of $ x_i$ and $ y_i$ is odd. So there are exactly $ n$ coordinates which are odd, which is a contradiction since we showed above that the number of odd coordinates is even. Hence $ d$ is even, so $ d^2\equiv 0\mod 4$. But we are also assuming that there is at least one odd coordinate, so for some $ i$, we have $ x_i^2 + y_i^2\equiv 1$ or $ 2\mod 4$, a contradiction. Note this generalizes to where $ d^2$ is any integer, not necessarily that $ d$ is an integer. There is some pair $ (x_i,y_i)$ with $ x_i,y_i$ odd, and there is some pair $ (x_j,y_j)$ (since otherwise every coordinate is odd), so $ x_i^2 + y_i^2\not\equiv x_j^2 + y_j^2\mod 4$.

the distance we move away from point = distance coming back therefore, total distace=d1+d2 ,, let they be expressed in no. of edges if d1is odd then d2 is odd so, sum is even and same in case of even edges . hence, no. of edges is even connecting firt two edges gives 2 vertices and +1 on every subsequent edge, but +-0 on the last edge hence, +2 on 1st edge and then no. of vertices willbe even on no. of edges as there will be +1 always until last edge . at that time, edge is odd so vertex is even and there will no addition of vertex on last edge. hence no. of vertices is even.so, this situation aint possible