a)
[tex]\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
A&({{ 4}}\quad ,&{{ 6}})\quad
% (c,d)
B&({{ 28}}\quad ,&{{ 11}})
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies
\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{11-6}{28-4}\implies \cfrac{5}{24}[/tex]
[tex]\bf y-{{ y_1}}={{ m}}(x-{{ x_1}})\implies y-6=\cfrac{5}{24}(x-4)\\
\left. \qquad \right. \uparrow\\
\textit{point-slope form}
\\\\\\
y-6=\cfrac{5}{24}x-\cfrac{5}{6}\implies y=\cfrac{5}{24}x-\cfrac{5}{6}+6
\\\\\\
y=\cfrac{5}{24}x+\cfrac{31}{6}[/tex]
b)
[tex]\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
A&({{ 4}}\quad ,&{{ 6}})\quad
% (c,d)
B&({{ 28}}\quad ,&{{ 11}})
\end{array}\qquad
% distance value
d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}
\\\\\\
AB=\sqrt{(28-4)^2+(11-6)^2}\implies AB=\sqrt{24^2+5^2}\implies AB=\sqrt{601}[/tex]
c)
well, the speed of the ship, the ship took two hours to cover the distance AB, thus the speed is length/time
[tex]\bf \cfrac{AB\ kms}{2hours}\implies \cfrac{\sqrt{601}\ kms}{2\ hrs}\approx 12.25765\ \frac{kms}{hrs}[/tex]
