Review question

# Can we show that $OAQY$ is a parallelogram? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R8704

## Solution

In the figure the points $X$ and $Y$ are such that $AX = \dfrac{1}{2}XB$ and $OY = YX$, while the point $P$ is such that $OA = 3AP$. The lines $YQ$ and $PQ$ are parallel to $OA$ and $OB$ respectively.

Given that $\mathbf{OA} = \mathbf{a}$ and $\mathbf{OB} = \mathbf{b}$, express $\mathbf{OP}$ and $\mathbf{OY}$ in terms of $\mathbf{a}$ and $\mathbf{b}$.

Since $\mathbf{OA} = 3\mathbf{AP}$, we know $\mathbf{OP}$ is $\dfrac{4}{3}\mathbf{a}$.

We can see that the vector $\mathbf{OX} = \mathbf{OA} + \mathbf{AX} = \mathbf{a} + \dfrac{1}{3}(\mathbf{b}-\mathbf{a}) = \dfrac{2}{3}\mathbf{a} + \dfrac{1}{3}\mathbf{b}.$

We also have that the vector $\mathbf{OY}=\dfrac{1}{2}\mathbf{OX} = \dfrac{1}{3}\mathbf{a} + \dfrac{1}{6}\mathbf{b}.$

Given that $\mathbf{YQ} = m\mathbf{a}$ and $\mathbf{PQ} = n\mathbf{b}$, find the values of $m$ and $n$. Hence show that $OAQY$ is a parallelogram.

The diagram tells us that $\mathbf{OP} + \mathbf{PQ} = \mathbf{OY} + \mathbf{YQ}$.

Thus $\dfrac{4}{3}\mathbf{a} + n\mathbf{b} = \dfrac{1}{3}\mathbf{a}+\dfrac{1}{6}\mathbf{b} + m\mathbf{a}.$

Since $\mathbf{a}$ and $\mathbf{b}$ are linearly independent (they are not parallel) we can equate coefficients of $\mathbf{a}$ and $\mathbf{b}$ on either side of this equation.

We are effectively saying here that if $\alpha \mathbf{a} + \beta \mathbf{b} = \gamma \mathbf{a} + \delta \mathbf{b},$ then $\alpha = \gamma, \beta = \delta.$ Is that always true?

This tells us $\dfrac{4}{3}= \dfrac{1}{3} + m \implies m = 1,$ and $n =\dfrac{1}{6}.$

Since $m = 1$, we have that $\big \vert \mathbf{AO} \big \vert= \big \vert\mathbf{QY}\big \vert$, and we know these two lines are parallel, so therefore $OAQY$ is a parallelogram.