OABD is a parallelogram. O is the origin. A is the point (-7, 3) and $\overrightarrow{AB}=\begin{pmatrix}4\\12\end{pmatrix}$. If $\overrightarrow{AC}=\frac{1}{4}\overrightarrow{AB}$.

1. Find the vector from $A$ to $C$ using the given fraction of $\overrightarrow{AB}$.

$\overrightarrow{AC}=\dfrac14\overrightarrow{AB}=\dfrac14\begin{pmatrix}4\\12\end{pmatrix}=\begin{pmatrix}1\\3\end{pmatrix}$

2. Add the position vector of $A$ to $\overrightarrow{AC}$ to find the position vector of $C$.

Since $O$ is the origin and $A=(-7,3)$,
$\overrightarrow{OA}=\begin{pmatrix}-7\\3\end{pmatrix}$.

$\overrightarrow{OC}=\overrightarrow{OA}+\overrightarrow{AC}=\begin{pmatrix}-7\\3\end{pmatrix}+\begin{pmatrix}1\\3\end{pmatrix}=\begin{pmatrix}-6\\6\end{pmatrix}$

1. Find the vector $\overrightarrow{OD}$ using the parallelogram property.

Since $OABD$ is a parallelogram, opposite sides are equal and parallel.

Hence
$\overrightarrow{OD}=\overrightarrow{AB}=\begin{pmatrix}4\\12\end{pmatrix}$ is a possible interpretation only if $OD$ corresponds to $AB$ in the vertex order.

However, from the naming $OABD$, the adjacent sides from $O$ are $\overrightarrow{OA}$ and $\overrightarrow{OD}$, with $\overrightarrow{AB}$ parallel and equal to $\overrightarrow{OD}$. Therefore
$\overrightarrow{OD}=\begin{pmatrix}4\\12\end{pmatrix}.$

2. Calculate the magnitude of $\overrightarrow{OD}$.

$\left|\overrightarrow{OD}\right|=\sqrt{4^2+12^2}$
$=\sqrt{16+144}$
$=\sqrt{160}=4\sqrt{10}$

1. Compare the bases of the two triangles on the same straight line $AB$.

Since $\overrightarrow{AC}=\dfrac14\overrightarrow{AB}$,
$AC=\dfrac14 AB$ and hence
$CB=AB-AC=\dfrac34 AB.$

2. Use equal heights to form the ratio of areas.

Triangles $OAC$ and $OBC$ have bases $AC$ and $CB$ on the same straight line $AB$, and they share the same perpendicular height from $O$ to $AB$.

Therefore,