Curve traced by all positions of a special point

Question

In the plane of given triangle $ABC$ if $P$ is a point satisfying geometric condition $[PAB]=[PAC]$ (where $[.]$ represents area enclosed). Then the curve traced by all such positions of $P$ divides the plane in $m$ parts then find $5m$. I got no clue how to start and why we are saying that it will form a curve?

Answer 1

Presumably, whoever devised the question described the locus of the point $P$ as a curve dividing the plane into some number of parts because it is, in fact, a curve dividing the plane into some number of parts. A hint to determining whether this is true or not is to observe that the area of triangle $PAC$ is half the length of $PA$ times the (perpendicular) distance of $C$ from the line of $PA$, and the area of triangle $PAB$ is half the length of $PA$ times the distance of $B$ from the same line. Thus, for the areas of the triangles to be the same, the line $PA$ must be equidistant from the points $B$ and $C$.