Given a triangle $ABC$ with unequal sides. $P$ is the set of all points which are equidistant from $B$ and $C$ and $Q$ is the set of all points which are equidistant from $AB$ and $AC$, then find $n(P\cap Q)$.
The points in set P lie on a line. Similarly the points in set Q also lie on a line. 2 lines intersect only once hence $n(P\cap Q)=1$ according to me. But the answer given is 2 and I can't figure out why. I need help figuring out the right answer.
The set of points equidistant from point $B$ and point $C$ is a line. But the set of points equidistant from line $AB$ and line $AC$ is not a line. It is the union of two lines: the angle bisectors of the two angles formed by $AB$ and $AC$.
So $P$ intersects each of the lines in $Q$ once, for a total of two intersections.
(We need to know that $AB \ne AC$, otherwise the triangle is isosceles, and the perpendicular bisector $P$ overlaps with one of the angle bisectors in $Q$.)