For $a$, $b$, $c$ in $\mathbb{R}^2$, I know that the set $$A=t_1a+t_2b+t_3c$$ with $0\leq t_1,t_2,t_3\leq 1$ and $1=t_1+t_2+t_3$ is a triangle with vertices $a$, $b$, $c$.
Could someone explain to me how we came up with those parameters geometrically? How could you explain it and why do they work?
Thank you in advance.
You forgot to add that $a,b,c\in\Bbb R^2$. Clearly, $a,b,c\in A$; for intance, $a=1\times a+0\times b+0\times c$. Furthermor, the line segment that goies from $a$ to $b$ is a subset of $A$, since any $p$ from that line segment can be written as $(1-t)a+tb$, for some $t\in[0,1]$, and therefore $p=(1-t)a+tb+0\times c$. For the same reason, $A$ contains the line segments joining $a$ and $c$ and joining $b$ and $c$.
Let $p$ be an element of $A$. If $t_1,t_2,t_3\in[0,1]$ are such that $t_1+t_2+t_3=0$, and if $p=t_1a+t_2+t_3c$, then\begin{align}p&=t_1a+t_2b+\bigl(1-(t_1+t_2)\bigr)c\\&=(t_1+t_2)\left(\frac{t_1}{t_1+t_2}a+\frac{t_2}{t_1+t_2}b\right)+\bigl(1-(t_1+t_2)\bigr)c.\end{align}Since$$\frac{t_1}{t_1+t_2}+\frac{t_2}{t_1+t_2}=1\quad\text{and}\quad\frac{t_1}{t_1+t_2},\frac{t_2}{t_1+t_2}\geqslant0,$$the point$$\frac{t_1}{t_1+t_2}a+\frac{t_2}{t_1+t_2}b\tag1$$belongs to the line segment joining $a$ to $b$, and therefore $p$ belongs the line segment joining $(1)$ to $c$; so, $p$ belongs to the triangle.
This works in the opposite direction too: if $p$ belongs to the trangle, find the point $q$ of the line segment joining $a$ to $b$ such that $q$ $p$ and $c$ are collinear. Then $q=(1-t)a+tb$, for some $t\in[0,1]$, and $p=(1-t')q+t'c$ for some $t'\in[0,1]$. So\begin{align}p&=(1-t')\bigl((1-t)a+tb\bigr)+t'c\\&=\bigl((1-t')(1-t)\bigr)a+(1-t')tb+t'c\\&\in A.\end{align}