If $C$ is on this line, can we find the ratio $AC:CB$?

If $C$ is on this line, can we find the ratio $AC:CB$?

Question

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 $AC:CB$.

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 $s$.

Suggestion

A diagram will help. How can we express the fact that $C$ is on $AB$?

If $AD$ is perpendicular to $DB$, what can we say about the gradients of $AD$ and $DB$?

Solution

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

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.

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.

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 $\mathbf{d}=\begin{pmatrix}2\\1\end{pmatrix}, s = 1.$