Review question

# What if we join the midpoints of opposite sides of a quadrilateral? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R8339

## Solution

The position vectors of four points $A, B, C, D$ relative to an origin $O$ are $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$, $\mathbf{d}$ respectively. Obtain the equation of the line $l$ joining the midpoints of $AB$ and $CD$.

The midpoint of $AB$ is $\dfrac{\mathbf{a}+\mathbf{b}}{2}$ while the midpoint of $CD$ is $\dfrac{\mathbf{c}+\mathbf{d}}{2}$. Using the standard form for a vector line $\mathbf{r} = \mathbf{u} + \lambda \mathbf{v},$ where $\mathbf{u}$ is the position vector of a point on the line and $\mathbf{v}$ is the direction of the line, we have that $l$ is the line $\mathbf{r} = \dfrac{\mathbf{a}+\mathbf{b}}{2} + \lambda \left(\left(\dfrac{\mathbf{c}+\mathbf{d}}{2}\right)-\left(\dfrac{\mathbf{a}+\mathbf{b}}{2} \right)\right),$ which becomes after simplification $\mathbf{r} = \dfrac{1-\lambda}{2}\mathbf{a}+ \dfrac{1-\lambda}{2}\mathbf{b}+ \dfrac{\lambda}{2}\mathbf{c}+ \dfrac{\lambda}{2}\mathbf{d}.$

Show that $l$ intersects the line joining the midpoints of $AC$ and $BD$ and find the position vector $\mathbf{p}$ of the point $P$ of intersection.

Deduce that the line joining the midpoints of $AD$ and $BC$ also passes through $P$.

If we do exactly the same for the line joining the midpoints of $AC$ and $BD$ (let’s call this $m$) using the parameter $\mu$ rather than $\lambda$, we arrive at the equation $\mathbf{r} = \dfrac{1-\mu}{2}\mathbf{a}+ \dfrac{\mu}{2}\mathbf{b} + \dfrac{1-\mu}{2}\mathbf{c} + \dfrac{\mu}{2}\mathbf{d}.$

Notice that if $\lambda = 0.5$, we get the point on $l$ given by $\mathbf{r} = \dfrac{1}{4}(\mathbf{a}+ \mathbf{b}+\mathbf{c}+\mathbf{d}).$

This is also the point on $m$ such that $\mu = 0.5$.

Thus the two lines $l$ and $m$ intersect at the point $P$ with position vector $\mathbf{p} = \dfrac{1}{4}(\mathbf{a}+ \mathbf{b} + \mathbf{c} + \mathbf{d})$.

We found the point $P$ by inspection, having noted the similarities between the two line equations.

The point $P$ is called the centroid of the four points $A$, $B$ , $C$ and $D$.

The above argument is completely symmetrical in $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$ and $\mathbf{d}$, which means that the line joining the midpoints of $AD$ and $BC$ must also pass through $P$.