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Thank you to Kristian from Maidstone Grammar School, Samuel from Wellington School and Niharika from Rugby School for submitting solutions to this problem.

Kristian and Samuel solved the questions by finding the equations of the lines first. Here is Kristian's solution to question 1.

To find the missing coordinate in any case, we need the equation of the line and this can be found using the equation
$$y-y_1=m(x-x_1)$$
We have $y_1$ and $x_1$, so we need $m$. We know that $m$ is the gradient so we know
$m=\frac{dy}{dx}=\frac{y_2-y_1}{x_2-x_1}$

$m=\frac{8-3}{8-2}=\frac{5}{6}$
Therefore
$y-3=\frac{5}{6}(x-2)$
$y=\frac{5x}{6}+\frac{4}{3}$

Now we know the line equation, we can just substitute the $x$ value into it

$y=\frac{5 \times 4}{6}-\frac{4}{3}=4.\dot{6}$


Niharika thought of some nice alternative methods for solving this problem which don't involve finding the equation of the line first:

The second way is by using the fact that the slope of a straight line is constant so
$\frac{8-a}{8-4}=\frac{8-3}{8-2}$
$8-a=\frac{10}{3}$
$a=4.\dot{6}$

The third way to do this would be to use similar triangles, which is the same as the second way.

Both methods can be applied to the remaining questions. Here are Samuel's answers to the remaining questions.

2. Work out gradient joining two known points, $\frac{5}{11}$.
Use $(y-y(1))=m(x-x(1))$
Using this, we derive
$11y=5x+63$
When $y=10$, $x=9.4$
Therefore $b=9.4$

3.i. Using the same equation as above, we derive that $4.8y=26x+71$
When $x=7.3$, $4.8y=260.8$
so $y=\frac{163}{3} = 54.\dot{3}$
Therefore $c=\frac{163}{3}$

3.ii. Using the same method as before, we derive
$y=\frac{28x}{5}-\frac{1057}{25}$
Solving for $x$ when $y=47.5$, $x=16.03214286$
Therefore $d=16.03$ (2d.p)

3.iii. Using the same method as before, we derive,
$y=-\frac{19x}{9}+\frac{332}{9}$
When $x=12$, $y=\frac{104}{9}$
Therefore, $e=\frac{104}{9}$

Here are Niharika's solutions using the similar triangles method.