Question: User: PQRS is a parallelogram, where P is the point $(1, 1)$
Given
$\vec{PQ} = \begin{pmatrix}3 \\ 6\end{pmatrix}$
$\vec{PS} = \begin{pmatrix}5 \\ 4\end{pmatrix}$
Determine the coordinates of $R$.
Graph: A parallelogram in the first quadrant with $P$ at $(1,1)$ near the origin, $Q$ above-right of $P$, $S$ farther right and slightly above $P$, and $R$ at the top-right; position_hint=center.
1. **State the problem:** We have a parallelogram $PQRS$ with point $P$ at coordinates $(1,1)$.
Vectors are given:
$$\vec{PQ} = \begin{pmatrix}3 \\ 6\end{pmatrix}, \quad \vec{PS} = \begin{pmatrix}5 \\ 4\end{pmatrix}$$
We need to find the coordinates of point $R$.
2. **Recall properties of parallelograms:**
In a parallelogram, the vector from $P$ to $R$ is the sum of vectors $\vec{PQ}$ and $\vec{PS}$:
$$\vec{PR} = \vec{PQ} + \vec{PS}$$
3. **Calculate $\vec{PR}$:**
$$\vec{PR} = \begin{pmatrix}3 \\ 6\end{pmatrix} + \begin{pmatrix}5 \\ 4\end{pmatrix} = \begin{pmatrix}3 + 5 \\ 6 + 4\end{pmatrix} = \begin{pmatrix}8 \\ 10\end{pmatrix}$$
4. **Find coordinates of $R$:**
Since $\vec{PR}$ is the vector from $P$ to $R$, and $P$ is at $(1,1)$, the coordinates of $R$ are:
$$R = P + \vec{PR} = \begin{pmatrix}1 \\ 1\end{pmatrix} + \begin{pmatrix}8 \\ 10\end{pmatrix} = \begin{pmatrix}1 + 8 \\ 1 + 10\end{pmatrix} = \begin{pmatrix}9 \\ 11\end{pmatrix}$$
**Final answer:** The coordinates of $R$ are $(9, 11)$.