Given a $\square$ $ABCD$ is it possible to see by intuition (no barycenter or scalar product) what this set represents : $\mathcal{E}:=\{M\ / \ \Vert 3\vec{MA}+\vec{MD}\Vert = \Vert 3\vec{MC}+\vec{MB}\Vert \}$ ?
Thanks in advance !
Given a $\square$ $ABCD$ is it possible to see by intuition (no barycenter or scalar product) what this set represents : $\mathcal{E}:=\{M\ / \ \Vert 3\vec{MA}+\vec{MD}\Vert = \Vert 3\vec{MC}+\vec{MB}\Vert \}$ ?
Thanks in advance !
On
If you rewrite the equation defining locus $\mathcal{E}$ like this: $$ (4M-3A-D)^2=(4M-3C-B)^2 $$ you can see that terms in $M^2$ cancel out, so the equation of the locus is linear in $M$, that is the locus is a straight line.
You only need then to find two points of $\mathcal{E}$: one of them is the square center, the other can be searched, for instance, on segment $AB$.
HINT
Consider the following figure-
Divide $DC$ so that you get $4$ intervals of equal length. Do the same with $AB$. The straight connecting the black points is the locus of the points the OP is looking for.