Scalar times Point + Scalar times Point? Let $P$, $Q$ be a pair of points in the Euclidean plane and let $t_1$, $t_2$ be a pair of scalars. My textbook says that the following operations are nonsense:
$$P + Q\\
t_1 \cdot P$$
However $t_1 \cdot P + t_2 \cdot Q$ is completely valid and makes sense! 
I cannot understand this at all. If you cannot multiply a scalar by a point and you cannot add points together, then how come that this expression is still valid? And how come that the result is a point?
 A: $t_1P+t_2Q$ only makes sense if $t_1+t_2=1$. The important thing here is that the operation is independent of the choice of coordinate frame. If $t_1+t_2=1$, the point is just an interpolation on the line between the two points. If not, you have a leftover term that shifts the point around depending on where your coordinate origin is.
The idea here is that a point and a vector are not the same. A vector is a difference (displacement) between two points and doesn't have an anchor in the coordinate plane. You can translate it around, multiply it by any scalar and add it to any point because it's a displacement. But a point is a physical place in the plane. You can't multiply a place by a scalar (what's New York times 2?). But you can multiply a displacement (2 steps north times 2 = 4 steps north).
An interpolation is ok, because you are essentially taking $D=Q-P$, the displacement vector from P to Q. Now start at $P$ and go a fraction of this distance:
$$R=P+t_1(Q-P)=(1-t_1)P+t_1 Q$$
Notice how the sum of the coefficients is $1$.

Formally, you can imagine adding a third component (use homogeneous coordinates). A point has a form $(x,y,1)$ and a displacement is $(x,y,0)$. If your operation sets the third component to $1$, the result is a point, if it is $0$, the result is a vector, and if it's anything else, the result is neither of those things.
