Relative to an origin \(O\), the position vectors of two points, \(A\) and \(B\), are \(\begin{pmatrix}2\\5\end{pmatrix}\) and \(\begin{pmatrix}4\\1\end{pmatrix}\) respectively. The point \(C\) is on \(AB\). Given that the position of \(C\) is \(\begin{pmatrix}2t\\t\end{pmatrix}\), find the ratio \(AC:CB\).

Intersection of the two lines

Since \(C\) lies on the line \(AB\), we can write its position vector as \[\lambda\begin{pmatrix}2\\5\end{pmatrix} + (1-\lambda)\begin{pmatrix}4\\1\end{pmatrix} =\begin{pmatrix}4-2\lambda\\4\lambda +1\end{pmatrix}.\]

Equating the coefficients, we have \(2t= 4-2\lambda\) and \(t = 4\lambda +1\). Solving these simultaneous equations gives \(t = \dfrac{9}{5}\) and \(\lambda = \dfrac{1}{5}\).

The ratio \(AC:CB = (1-\lambda):\lambda = 4:1\).

The point \(D\) is on \(OC\), between \(O\) and \(C\), and \(\widehat{ADB}\) is a right angle. Given that the position vector of \(D\) is \(\begin{pmatrix}2s\\s\end{pmatrix}\), find the value of \(s\).

Approach 1: gradients multiply to \(-1\)

We can see that the gradient of \(AD\) is \(\dfrac{5-s}{2-2s}\), while the gradient of \(DB\) is \(\dfrac{1-s}{4-2s}\), and since \(AD\) and \(BD\) are perpendicular,

\[\dfrac{5-s}{2-2s}\times\dfrac{1-s}{4-2s} = -1\]

which gives us that \[5s^2-18s+13=0.\]

This factorises into \[(5s-13)(s-1) = 0 \implies s = 1, \dfrac{13}{5}.\]

The value \(\dfrac{13}{5}\) gives \(D\) to be further from the origin than \(C\), so we can discard this, and \[D=(2,1).\]

The value of \(s\) we seek is \(1\).

Approach 2: using the angle in a semicircle

If \(A\),\(B\) and \(D\) are points on a circle with \(AB\) a diameter of that circle, then \(\widehat{ADB}\) is a right angle.

The converse is also true; if \(\widehat{ADB}=\dfrac{\pi}{2}\), then \(AB\) has to be a diameter of the circle.

the angle in a semicircle

Thus \(D\) is on the circle with diameter \(AB\). The midpoint of \(AB\) is \((3,3)\) and the length of \(AB\) is \(2\sqrt{5}\), and so the circle has centre \((3,3)\) and radius \(\sqrt{5}\).

Thus the equation of the circle is \((x-3)^2+(y-3)^2=5\). The point \(D\) is on this, so \((2s-3)^2 +(s-3)^2 = 5\), or \(5s^2-18s+13=0\).

The solution then proceeds as before.

Intersection of circle and line

Approach 3: using the scalar product

We know that the lines \(DA\) and \(DB\) have to be perpendicular.

Thus \((\mathbf{d}-\mathbf{a})\cdot(\mathbf{d}-\mathbf{b})=0,\) where \(\mathbf{d}\) is the position vector of \(D\), and where \(\cdot\) represents the scalar product.

Now \(\mathbf{d} = \begin{pmatrix}2s\\s\end{pmatrix},\) so \(\mathbf{d}-\mathbf{a} =\begin{pmatrix}2s-2\\s-5\end{pmatrix} , \mathbf{d}-\mathbf{b}= \begin{pmatrix}2s-4\\s-1\end{pmatrix}.\)

Thus we have \[\begin{align*} (\mathbf{d}-\mathbf{a})\cdot(\mathbf{d}-\mathbf{b})&=\begin{pmatrix}2s-2\\s-5\end{pmatrix} \cdot \begin{pmatrix}2s-4\\s-1\end{pmatrix}\\ &=(2s-2)(2s-4)+(s-5)(s-1)\\ &=5s^2-18s+13=0, \end{align*}\]

as before, and so \[\mathbf{d}=\begin{pmatrix}2\\1\end{pmatrix}, s = 1.\]

The line AB with the point D marked so that the angle ADB is a right angle