Linear combinations with restrictions on coefficients? I understand why all linear combinations of two linearly independent vectors $u, v \in \Bbb R^2$ represent the entire plane, but I'm having difficulty interpreting the case where different restrictions are placed on the coefficients. For example, the general form of a linear combination is given below
$$ c \cdot u + d \cdot v $$
where $c, d \in \Bbb R$
Now what happens if we impose the restriction $c + d = 1$? I believe this will represent a line, but I don't understand why. 
As another example consider the diagram below
$\hspace{6cm}$
According to my textbook, you can fill in the dashed triangle with the following two restrictions
$1.) \hspace{2mm} c, d, e \geq 0$
$2.) \hspace{2mm} c + d + e = 1$
But again, I'm not sure why
Question: Why does the restriction $c + d = 1$ generate a line and the other two restrictions fill the dashed triangle?
Any other notes on interpreting restrictions on coefficients would also be appreciated
 A: For the restriction $c+d=1$, rewrite this as
$$cu+(1-c)v=v+c(u-v)$$
 For $0\leq c$ this is a ray starting at $v$ and pointing towards $u$, as when $c=0$ the value is $v$ and we translate the vector in the direction of $u-v$ $1$ unit until we reach $u$, then continue to infinity. For $c<0$ we have a ray pointing in the opposite direction.
For the restriction $c,d,e\geq 0$, $c+d+e=1$, write this as
$$cu+dv+(1-c-d)w=w+c(u-w)+d(v-w)$$
Intuitively this is the same, but we are translating in two directions starting at $w$, fanning out in one direction towards $u$ and another towards $v$. Ranging over arbitrary positive values of $c,d$ this is the area in the plane through $w$ spanned by $u-w,v-w$ between the two rays from $w$ to $u$ and $w$ to $v$. We obviously do not get this entire area because $c+d\leq 1$. Because of this condition there is symmetry here. We can rewrite it as
$$v+c(u-v)+e(w-v)$$
and
$$u+d(v-u)+e(w-u)$$
Thus the area is in fact the triangle enclosed by the three lines through the points in the unique plane containing them (assuming nondegeneracy).
A: the condition $c+d=1$ is easy.
take $$\vec{x}=c\vec{u}+d\vec{v}$$
now substitute $d\to1-c$, you obtain:
$$
\vec{x}=c(\vec{u}-\vec{v})+\vec{v}\longrightarrow\lambda\,\vec{v_0}+\vec{v_1} \textrm{ (vectorial line equation)}
$$
if you vary $c$ continously you obtain a line.
for the other one you should look into Convex Combination (the explanation is longer and a rather tedious one).
