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**earboth** 1. Use a coordinate system, place the origin at the intersection. Then the startpoint is S(2,0) and the endpoint is E(0,3).

2. The direction vector of Camerons way is: $\displaystyle (0,3) - (2,0) = (-2,3)$

3. The vector equation, which describes Camerons way, is:

$\displaystyle (x,y)=(2,0)+t\cdot(-2,3)~\implies~\boxed{\begin{array}{l}x=2-2t\\y=3t\end{array}}$

4. The distance between any point of this line and the origin(= intersection) is calculated by:

$\displaystyle d(t) = \sqrt{ (2-2t)^2+(3t)^2} = \sqrt{13t^2-4t+4}$

5. Calculate the first derivative of d:

$\displaystyle d'(t)=\frac12 \cdot (13t^2-4t+4)^{-\frac12} \cdot (26t-4) = \dfrac{13t-2}{\sqrt{13t^2-4t+4}}$

6. Now plug in $\displaystyle t=\frac14$ to calculate the rate of change. I've got $\displaystyle d'\left(\frac14\right)=\frac5{61} \sqrt{61} \approx 0.6402$