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Click here for the question.

I worked out that $m_{AB} = -2$,and therefore $m_{DCE} = \frac{1}{2}$.

The midpoint of $AB$(which is $C$) is $(2,4)$, then I substituted $(2,4)$ into $y = \frac{1}{2}x +c$
which gave me $y = \frac{1}{2}x +3$ as the equation of the line $DCE$.

I checked the mark scheme and their answer is $y = \frac{1}{2}x -2$. I was doing another of their questions and the mark scheme was definitely wrong then.

Was my method/answer correct, if not, where did I go wrong?

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The rectangular coordinates of the midpoint $C$ of line segment $ \overline{AB}$ are calculated traditionally as $\left(\frac{0+4}{2}, \frac{8+0}{2} \right) = (2,4)$. Also, the slope of said line segment is $\frac{8 - 0}{0 - 4} = -2 := m$.

Therefore, the line $\ell$ through $D$ and $E$ has slope $-\frac{1}{m} = \frac{1}{2}$ and goes through the point $C = (2,4)$.

From this you should easily be able to find the equation of $\ell$.

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    I did that (see above), it was when it came to substituting that I got the y-intercept wrong in the answer and I'm not sure how. – 2017-02-12
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    Sorry about that. Admittedly your notation wasn't clear to me. If $y = \frac{1}{2}x + c$ is the equation for $\ell$, then you should have calculated $c$ by setting $4 = \frac{1}{2}(2) + c \Longrightarrow c = 3$. So it seems to me that you were correct all along. Perhaps the author gave the incorrect answer. The line should be $y = \frac{1}{2} x + 3$ – 2017-02-12
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    No worries and apologies about the notation. Thank you for helping, much appreciated! :) – 2017-02-12