Solution:

For simplicity we can find the adjacency matrix which gives distances of all object from each other. Using Euclidean distance we have

$\begin{aligned} \mathrm{D}(\mathrm{i}, \mathrm{j}) &=\sqrt{\left|\mathrm{x}_{2}-\mathrm{x}_{1}\right|^{2}+\left|\mathrm{y}_{2}-\mathrm{y}_{1}\right|^{2}} \\ \mathrm{D}(\mathrm{A}, \mathrm{B}) &=\sqrt{(2-3)^{2}+(2-2)^{2}}=1 \end{aligned}$

Similarly we can compute for the rest.

(i) Singlelink:

step 1: Since $C, E$ is minimum we can combine clusters $C, E$

Step 2: Now $\mathrm{A}$ and $\mathrm{B}$ is having minimum value therefore we merge these two clusters.

Step 3 : Cluster $(\mathrm{A}, \mathrm{B})$ and $\mathrm{D}$ can be merged together as they are having minimum distance value

Step 4 : In the last step there are only two clusters to be combined they are, $(\mathrm{A}, \mathrm{B}, \mathrm{D})$ and $(\mathrm{C}, \mathrm{E})$

Now the final dendrogram is