The slope formula for tow given points is
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]where, in our case,
[tex]\begin{gathered} (x_1,y_1)=(\frac{2}{5},\frac{14}{21}) \\ (x_2,y_2)=(\frac{1}{3},\frac{4}{12}) \end{gathered}[/tex]By substituting these values into our formula, we have
[tex]m=\frac{\frac{4}{12}-\frac{14}{21}}{\frac{1}{3}-\frac{2}{5}}[/tex]Lets compute the numerator first. We have
[tex]\frac{4}{12}-\frac{14}{21}=\frac{4\cdot21-12\cdot14}{12\cdot21}[/tex]which gives
[tex]\frac{4}{12}-\frac{14}{21}=\frac{84-168}{12\cdot21}=\frac{-84}{252}=-\frac{7}{21}[/tex]Similarly, in the denominator we have
[tex]\frac{1}{3}-\frac{2}{5}=\frac{5\cdot1-2\cdot3}{3\cdot5}=\frac{5-6}{15}=-\frac{1}{15}[/tex]Then, our slope is given by
[tex]m=\frac{-\frac{7}{21}}{-\frac{1}{15}}[/tex]By applying the sandwhich law, we get
then, the slope is m=5. Then, our line has the followin form
[tex]y=mx+b\Rightarrow y=5x+b[/tex]where b is the y-intercept. We can find b by substituting one of the 2 given points, that is, if we substitute point (1/3,4,12) into the last expression, we have
[tex]\frac{4}{12}=5(\frac{1}{3})+b[/tex]which gives
[tex]\frac{1}{3}=\frac{5}{3}+b[/tex]then b is given by
[tex]\begin{gathered} b=\frac{1}{3}-\frac{5}{3} \\ b=\frac{1-5}{3} \\ b=-\frac{4}{3} \end{gathered}[/tex]And finally, the line equation in slope-intercept form is
[tex]y=5x-\frac{4}{3}[/tex]
