How to show that the trianglular plane with given vertices is a close set? Let Δ denote the sides along with the "inside" of the triangle whose vertices are at (−1,0), (1,0) and (0,1). Show that Δ is closed.
I just have one idea to do this. If P be any point in the triangle ABC then area PAB + area PBC + area PAC= area ABC. Since the vertices are give then we can calculate this area and we will choose P=(x,y). Then we will define a set
E={p=(x,y)∈R2:areaPAB+areaPBC+areaPAC=areaABC}.
After defining I will try to show that the complement of this set is open. But if I go through this steps then the sum becomes so complex. I just want to know if there any simpler idea how to solve this. Because in previous exercise I have showed any plane P in R3 is closed. And this is also a triangular plane. I want help.

First note that for any reals $a$ and $b$, which are not both zero, any halfplane $H=\{(x,y)\in\Bbb R^2: ax+by+c\ge 0\}$ is a closed subset of a plane $\Bbb R^2$ endowed with the usual topology, because $H$  is a preimage of a closed subset $[0,\infty)$ of $\Bbb R$ under a continuous map from $\Bbb R^2\to \Bbb R$, $(x,y)\mapsto ax+by+c$. The triangle is closed, being an intersection of three such halfplanes
