ABCD 是平行四边形。A 是点 $(-6,-2)$，B 是点 $(-4,-7)$。$\overrightarrow{BC} = \begin{pmatrix}8\\-2\end{pmatrix}$。

1. Find the horizontal and vertical changes from A to B.

For $A(-6,-2)$ and $B(-4,-7)$,

$\Delta x=-4-(-6)=2$

$\Delta y=-7-(-2)=-5$

2. Apply the distance formula to find the length of AB.

$AB=\sqrt{(2)^2+(-5)^2}$

$AB=\sqrt{4+25}=\sqrt{29}$

1. Find the coordinates of $C$ using the vector $\overrightarrow{BC}$.

Since $B=(-4,-7)$ and $\overrightarrow{BC}=\begin{pmatrix}8\\-2\end{pmatrix}$,

$C=(-4+8,\,-7-2)=(4,-9).$

2. Find the gradient of $CD$ using the fact that $CD$ is parallel to $AB$.

In a parallelogram, opposite sides are parallel, so $CD \parallel AB$.

For $A=(-6,-2)$ and $B=(-4,-7)$,

$m_{AB}=\frac{-7-(-2)}{-4-(-6)}=\frac{-5}{2}.$

Hence,

$m_{CD}=-\frac{5}{2}.$

3. Form the equation of line $CD$ using its gradient and point $C$.

Using $C=(4,-9)$ and gradient $-\dfrac{5}{2}$,

$y-(-9)=-\frac{5}{2}(x-4).$

$y+9=-\frac{5}{2}x+10$

$y=-\frac{5}{2}x+1.$

1. Find the vector $\overrightarrow{AC}$ using the coordinates of A and C.

From earlier, $C=(4,-9)$ and $A=(-6,-2)$.

$\overrightarrow{AC}=\begin{pmatrix}4-(-6)\\-9-(-2)\end{pmatrix}=\begin{pmatrix}10\\-7\end{pmatrix}$

2. Use the diagonals-bisect property of a parallelogram to find $\overrightarrow{XC}$.

In a parallelogram, the diagonals bisect each other, so X is the midpoint of AC.

Therefore,
$\overrightarrow{XC}=\frac{1}{2}\overrightarrow{AC}$

$\overrightarrow{XC}=\frac{1}{2}\begin{pmatrix}10\\-7\end{pmatrix}=\begin{pmatrix}5\\-\dfrac{7}{2}\end{pmatrix}$

1. Find the coordinates, hence position vector, of $C$.

From $B(-4,-7)$ and $\overrightarrow{BC}=\begin{pmatrix}8\\-2\end{pmatrix}$,
$C=(-4+8,\,-7-2)=(4,-9).$
So
$\overrightarrow{OC}=\begin{pmatrix}4\\-9\end{pmatrix}.$

2. Use $\overrightarrow{XC}$ to find the position vector of $X$.

From part (c)(i),
$\overrightarrow{XC}=\begin{pmatrix}5\\-\dfrac{7}{2}\end{pmatrix}.$
Since $\overrightarrow{OC}=\overrightarrow{OX}+\overrightarrow{XC}$,
$\overrightarrow{OX}=\overrightarrow{OC}-\overrightarrow{XC}.$

1. Relate the ratio of the triangle areas to the ratio of the bases on $BC$.

Since $A$ is the common vertex of $\triangle ABP$ and $\triangle ACP$, and $B$, $P$, $C$ lie on the same straight line, both triangles have the same perpendicular height from $A$ to $BC$.

Hence
$\text{area of }\triangle ABP : \text{area of }\triangle ACP = BP : PC = 2:1.$