Finding the second point of a vector knowing the magnitude

Question

Points $A$ and $B$ lie in a plane, and $A$ has coordinates $(6, -2)$. Find four possible positions for $B$ such that $\lVert AB\rVert = \sqrt{13}$

How do I apply Pythagoras Theorem to this?

Answer 1

To find whole-number solutions, you look for integers $m$ and $n$ such that $m^2+n^2=13$. This shouldn't take you too long. This gives you eight suitable points (assuming $m\ne n$): $(6\pm m, -2\pm n)$ and $(6\pm n, -2\pm m)$.

If this is not clear to you, just construct a couple of those points and compute their distance to $A$, and you will see how it works.