By applying the knowledge of coordinate geometry,
[tex]\begin{gathered} \text{Two points on the graph = (5, 130) and (3, 100)} \\ \text{Gradient of the graph, m = }\frac{y_2-y_1}{t_2-t_1}=\text{ }\frac{130-100}{5-3}=\text{ }\frac{30}{2}=15 \\ \text{Equation of the line = }y-y_1=m(t-t_{1)}where(t_1,y_1)=(3,100) \\ =\text{ y-}100\text{ = 15(t -3)} \\ y\text{ =}15t-45+100 \\ y=\text{ 15t + }55 \end{gathered}[/tex]
