Does the line $(1, 2) + t (1, 1)$ contain the point $(3,3)$? How would I determine if a line contains a point?
ie.
$[1, 2] + T [ 1, 1]$
with a point of $[3, 3]$
I know how to do this problem if the vector was in a different form, but the T portion of it confuses me slightly.
I'm assuming that I would try to set $T$ to a scalar such that it would equal $[3, 3]$, which would be impossible in this case, thus the point is not in the line?
 A: A line is a collection of points following certain rules. In your question, this collection of points was given in terms of a starting point ($[1,2]$), and a direction to move to arrive at new points ($[1,1]$). You can visualize the variable $T$ as time: after two units of time ($T=2$), you have moved two units up and two units to the right from your starting point ($2*[1,1]=[2,2]$). Now, we take into consideration our starting point by adding it to how far we have moved ($[1,2]+[2,2]=[3,4]$). So, after walking from $[1,2]$ in direction (and speed, but that can be ignored for intuition) $[1,1]$ for $2$ minutes, we arrive at $[3,4]$.
If you want to find out if a line contains a point, you can use the following technique. Consider that your line starts at $[1,2]$ and the only possible way there is to move is some multiple of up one unit and right one unit, or $T*[1,1]$. In order to see if we arrive at $[3,3]$ at some point in time on our infinitely long walk, we want to have $[1,2]+T*[1,1]=[3,3]$. This gives us two equations which must be simultaneously satisfied in order for $[3,3]$ to be on the line: $1+T*1=3$ and $2+T*1=3$. You can use whatever technique you are comfortable with to solve this system of equations. In the end, it can be seen that there is no such $T$.
A: The set of points on a line can be represented in various ways. One way is to specify it as
$$
\{[x_0,y_0]+t[u,v]:t\in\mathbb{R}\}
$$
where $[x_0,y_0]$ and $[u,v]$ are vectors, with $[u,v]\ne[0,0]$. This means that a point $[x,y]$ belongs to the line if and only if there exists $t\in\mathbb{R}$ with $[x,y]=[x_0,y_0]+t[u,v]$. This is called a parametric representation.
Another way is to specify it as
$$
\{[x,y]:ax+by+c=0\}
$$
for some $a,b,c$ with $a$ and $b$ not both zero. This is the most common way, but the parametric representation has its uses.
How can we go from one representation to the other? Suppose $[x,y]=[x_0,y_0]+t[u,v]$. Then
$$
\begin{cases}
tu=x-x_0 \\[4px]
tv=y-y_0
\end{cases}
$$
If $v\ne0$, we have $t=(y-y_0)/v$ and so $u(y-y_0)=v(x-x_0)$. Conversely, any point $[x,y]$ satisfying $vx-uy-(vx_0-uy_0)=0$ can be represented as
$$
[x,y]=[x_0,y_0]+\frac{y-y_0}{v}[u,v]
$$
If $v=0$, then $u\ne0$ and so we get the points of the line $y-y_0=0$.
Conversely, given the line with equation $ax+by+c=0$, we can consider the vector $[b,-a]$ and an arbitrary point $[x_0,y_0]$ on the line, finding that every point on the line can be represented as
$$
[x,y]=[x_0,y_0]+t[b,-a]
$$
Indeed, from $ax+by+c=0$ and $ax_0+by_0+c=0$ we derive $a(x-x_0)+b(y-y_0)=0$; finding the expression for $t$ is easy.
Note that there's no uniqueness in both cases. In the first representation the “base point” $[x_0,y_0]$ is an arbitrary point of the line and $[u,v]$ is determined up to a nonzero scalar factor (it is any nonzero vector orthogonal to the line). In the second representation the coefficients $a$, $b$ and $c$ are determined up to a nonzero scalar factor.
In your case the line is, in the second form, $1\cdot(x-1)=1\cdot(y-2)$ or, simplifying, $x-y+1=0$. The point $[3,3]$ doesn't satisfy this equation.
You don't really need to write the equation: there is no $t$ such that
$$
[3,3]=[1,2]+t[1,1]
$$
because this would mean $t[1,1]=[2,1]$.
A: We would need that:
\begin{align}
x: 3=1+t\\
y: 3=2+t
\end{align}
Subtracting the equations, we get that $0=1$, which is impossible.
So $(3,3)$ is NOT on the line $(1,2)+t(1,1)$.
