First, we can use the points (4,3) and (20,15) to find the slope of the line:
[tex]\begin{gathered} s=\frac{m_2─m_1}{j_2─j_1}=\frac{15─3}{20─4}=\frac{12}{16}=\frac{3}{4} \\ s=\frac{3}{4} \end{gathered}[/tex]Now we can find the relation function using the point (4,3) and the slope s=3/4
[tex]\begin{gathered} m─m_1=s(j─j_1) \\ \Rightarrow m─3=\frac{3}{4}(j─4)=\frac{3}{4}j─(\frac{12}{4})=\frac{3}{4}j─3 \\ m=\frac{3}{4}j \end{gathered}[/tex]therefore, the equation to describe the relationship between J and M is m=3/4j
