Point $A, B, C, D$ have position vectors $a, b, c, d$ respectively relative to an origin O.
If $P$ divides $AB$ in the ratio $1:2$ and $Q$ divides $CD$ in the ratio $1:2$, obtain an expression for the position vector of $X$, where $X$ is the midpoint of $PQ$.
If $ABCD$ is a parallelogram show that $X$ is the point in which the diagonals $AC$ and $BD$ intersect.
I did the following diagram to work out this problem.
I figured that to get $\overrightarrow{OX}$, I need $\overrightarrow{PQ}$, $\overrightarrow{PX}$ and $\overrightarrow{AP}$, which I got as below,
$$
\begin{align}
\overrightarrow{PQ} &= \overrightarrow{PA} + \overrightarrow{AD} + \overrightarrow{DQ} \\
&= -\frac{2}{3} a – \frac{1}{3} b + \frac{2}{3} c + \frac{1}{3} d \\
\\
\overrightarrow{PX} &= \frac{1}{2}\overrightarrow{PQ} \\
&= -\frac{1}{3} a – \frac{1}{6} b + \frac{1}{3} c + \frac{1}{6} d \\
\\
\overrightarrow{AP} + \overrightarrow{PX} &= \overrightarrow{AX} \\
\overrightarrow{AX} &= \frac{1}{3} \overrightarrow{AB} + \overrightarrow{PX} \\
&= -\frac{2}{3}a + \frac{1}{6}b + \frac{1}{3}c + \frac{1}{6}d \\
\\
\overrightarrow{OX} &= \overrightarrow{OA} + \overrightarrow{AX} \\
&= \frac{1}{3}a + \frac{1}{6}b + \frac{1}{3}c + \frac{1}{6}d \\
&= \frac16(2a+b+2c+d)
\end{align}
$$
For the second part I figure I need to show that $X$ lies on $AC$ and $BD$. ie:- $\overrightarrow{AX} = k\overrightarrow{AC} |k\overrightarrow{XC}$ and $\overrightarrow{BX} = k\overrightarrow{BC} |k\overrightarrow{XD}$.
I tried doing this but end up with vectors without all the components. So I am unable to represent the vectors as multiples of each other. What am I missing? How do you solve the last part of this question?
Thanks for all your help!
Best Answer
In a parallelogram, you have $b-a=c-d$, so $b-a+d-c=0$. Adding one twelfth of that to your last displayed equation yields $\overrightarrow{OX}=\frac14(a+b+c+d)$, which is the midpoint of the parallelogram and the point where the diagonals intersect.