How to translate a vector and then rotate by a point I am trying to do this problem:

Identify the combination formed by first translating by the vector
  $(2,0)$ and then rotating by $90$ degrees about $(0,0)$.

but I'm a bit confused so,
I searched online where it said:

For the rotation,by an angle
  $\theta$ about point $P$. This can be replaced by two reflections by
  lines at an angle $\theta/2$ which intersect at P. 
  The translation by a distance $t$ can be replaced by reflections by two
  parallel lines at a distance of $t/2$, where the lines are
  perpendicular to the direction of the translation.
The second line of the rotation can be made equal to the first line of
  the translation. Thus they cancel in the composition, leaving only the
  first line of the rotation and the second line of the translation.
  This is a rotation through the point of intersection of these two
  lines.

However, as I was reading online, I read how translating and rotating is not commutative. Therefore, the above procedure only helps me when I rotate first then translate next. So what happens when I translate a vector then rotate as in the problem mentioned above.
 A: If $T$ denotes your translation by $(2,0)$, and $R$ your rotation by $90^\circ$ around the origin, then your map is $Q=R\circ T$. You can find $Q$ either analytically or geometrically.
Analytic solution 
$T$ and $R$ are defined by
$$
T(x)=x+(2,0)=(x_1+2,x_2), \, R(x)=(-x_2,x_1) \quad \forall x \in \mathbb{R}^2.
$$
Therefore
$$
Q(x)=R(T(x))=R(x_1+2,x_2)=(-x_2,x_1+2) \quad \forall x=(x_1,x_2)\in \mathbb{R}^2
$$
Notice that $Q$ has exactly one fixed point, i.e. $a=(-1,1)$. In fact $Q(x)=x \iff x=(-1,1)$ and therefore $Q$ is a rotation.
For every $x\in \mathbb{R}^2$ we have
$$
(x-a)\cdot(Q(x)-a)=(x_1+1,x_2-1)\cdot(-x_2+1,x_1+1)=0,
$$
i.e. $(x-a)\perp (Q(x)-a)$, and therefore $Q$ is a rotation by $90^\circ$ around the point $a=(-1,1)$.
Geometric solution
If $\Delta\subset \mathbb{R}^2$ is a straight line, we denote by $S_\Delta$ the reflection about $\Delta$.  We can then write
$$
T=S_{L_0}\circ S_{L_1}, \quad R=S_{L_2}\circ S_{L_0},
$$
with
\begin{eqnarray}
L_0&=&\{(x_1,x_2)\in \mathbb{R}^2:\, x_1=0 \},\\ 
L_1&=&\{(x_1,x_2)\in \mathbb{R}^2:\, x_1=1 \},\\
L_2&=&\{(x_1,x_2)\in \mathbb{R}^2:\, x_2=-x_1 \}
\end{eqnarray}
We have
$$
Q=R\circ T=(S_{L_2}\circ S_{L_0})\circ(S_{L_0}\circ S_{L_1})=S_{L_2}\circ (\underbrace{S_{L_0}\circ S_{L_0}}_{\mbox{identity}})\circ S_{L_1}=S_{L_2}\circ S_{L_1}
$$
Since the two lines $L_1$ and $L_2$ intersect at the point $a=(1,-1)$ and the angle $(L_1,L_2)$ between $L_1$ and $L_2$ is $45^\circ$, it follows that $Q$ is a rotation by $2\times45^\circ=90^\circ$ around the point $a=(-1,1)$.
A: Hint: Matrix Multiplication (not commutative!) Indeed, all isometries can be represented as 3 reflections, but not all isometries are rotations, translations, and reflections.
