CDEF is a quadrilateral - Edexcel - GCSE Maths - Question 1 - 2019 - Paper 2

Given that $M$ is the midpoint of $DE$ and $X$ is the point on $FM$ such that $FX : XM = n:1$ and $CVE$ is a straight line:

1. Since $M$ is the midpoint, we can express: $\vec{M} = \frac{\vec{D} + \vec{E}}{2}.$ Substituting for $\vec{D}$ and $\vec{E}$:

   $\vec{M} = \frac{\left(\vec{C} + \mathbf{a}\right) + \left(\vec{C} + \mathbf{a} + \mathbf{b}\right)}{2} = \vec{C} + \mathbf{a} + \frac{\mathbf{b}}{2}.$

2. Now, we can express $X$ on line $FM$:

   $\vec{X} = \frac{n}{n + 1} \vec{F} + \frac{1}{n + 1} \vec{M}.$

3. Using the conditions for collinearity of $C, V, E$, we can find $n$. By equating the ratios of segments, we derive:

   $n + 1 = 2 \text{ (from balance of weights)}$, yielding: $n = 1.$