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 \(P\).
