The gradient between the two points is \(\frac{1}{3}\). Hence, the perpendiular gradient is \(-\frac{3}{1}=-3\).

Midpoint: \(\left(\frac{-2+4}{2},\frac{5+3}{2}\right)=\left(1,4\right)\)

Gradient: \(\frac{3-5}{4--2}=-\frac{2}{6}=-\frac{1}{3}\)

Perpendicular Gradient: \(\frac{3}{1}=3\)

Equation: \(y=3x+c\)

Substitute \((1,4)\) into the equation: \(4=3\times1+c\), so \(c=1\).

Thus, \(y=3x+1\)

A vertical line which passes through \(4\) on the x-axis has equation, \(x =4\).

If the gradient is \(3\), we can write the gradient as \(\frac{3}{1}\) so that the perpendicular gradient is \(-\frac{1}{3}\).

The missing edge seperates sites A(1,4) and B(6,6).

Midpoint: \(\left(\frac{1+6}{2},\frac{4+6}{2}\right)=\left(3.5,5\right)\)

Gradient: \(\frac{6-4}{6-1}=\frac{2}{5}\)

Perpendicular Gradient: \(-\frac{5}{2}\)

Equation: \(y=-\frac{5}{2}x+c\)

Substitute \((3.5,5)\) into the equation: \(5=3.5\times-\frac{5}{2}+c\), so \(c=13.75\).

Thus, \(y=-\frac{5}{2}x+13.75\)

Since \(a,b,c \) must be integers with \(a,b>0\), we can multiply te equation by \(4\) and rearrange to obtain

\(10x+4y-55=0\)

The point (2, 3) is located on an edge between the sites C and D. We therefore take the mean of the rainfalls at C and D: \(\frac{18+20}{2}=19\ cm\)

If the point was located inside a cell, then we would estimate the rainfall to be the same as the site within that cell.

The point (5, 3) is located on a vertex between the sites B, C and D. We therefore take the mean of the rainfalls at B, C and D: \(\frac{22+18+20}{3}=20\ cm\)

When adding a new site, a perpendicular bistector needs to be found between te new site, and the site which already exists in the cell. Then a new perpendicular bistector must be found between the new site, and any sites in cells which are directly next to the new site.

The missing site is contained in the cell which currently does not have a site.

If we start by looking at the edge between the missing site and site D, it has gradient of 3 right and 1 down, \(-\frac{1}{3}\). Therefore the perpendicular gradient must be \(3\). Utilising the same idea with the edge between site B and the missing site, the gradient is \(-\frac{1}{2}\), which means the perpendicular gradient must be \(2\). Starting at site D and B, and drawing a line backwards with the perpendicular gradients found will ascertain the location of the missing site.

Since there are two vertices, we will need to check the distance between each vertex and a randomly picked site that each vertex seperates.

For the vertex \((9,5)\), if I randomly pick site A, the distance is given by, \(d=\sqrt{(9-13)^2+(5-4)^2}=\sqrt{(-4)^2+1^2}=\sqrt{17} \)

For the vertex (5, 6), if I randomly pick site D, the distance is given by,

\(d=\sqrt{(5-0)^2+(6-9)^2}=\sqrt{5^2+3^2}=\sqrt{34} \)

Therefore, the largest empty circle has centre \((5, 6)\) and radius \(\sqrt{34} \).