Answer: The required equation of the line in slope-intercept form is [tex]y=\dfrac{1}{2}x+\dfrac{3}{4}.[/tex]
Step-by-step explanation: We are given to find the equation of the line passing through the points [tex]\left(\dfrac{2}{5},\dfrac{19}{20}\right)[/tex] and [tex]\left(\dfrac{1}{3},\dfrac{11}{12}\right)[/tex] in slope-intercept form.
We know that
the slope of a line passing through the points (a, b) and (c, d) is given by
[tex]m=\dfrac{d-b}{c-a}.[/tex]
So, the slope of the given line will be
[tex]m=\dfrac{\frac{11}{12}-\frac{19}{20}}{\frac{1}{3}-\frac{2}{5}}=\dfrac{\frac{55-57}{60}}{\frac{5-6}{15}}=\dfrac{-2}{60}\times\dfrac{15}{-1}=\dfrac{1}{2}.[/tex]
Also, since the line passes through the point [tex]\left(\dfrac{2}{5},\dfrac{19}{20}\right)[/tex], so its equation will be
[tex]y-\dfrac{11}{12}=m\left(x-\dfrac{1}{3}\right)\\\\\\\Rightarrow y-\dfrac{11}{12}=\dfrac{1}{2}\left(x-\dfrac{1}{3}\right)\\\\\\\Rightarrow y=\dfrac{1}{2}x-\dfrac{1}{6}+\dfrac{11}{12}\\\\\\\Rightarrow y=\dfrac{1}{2}x+\dfrac{11-2}{12}\\\\\\\Rightarrow y=\dfrac{1}{2}x+\dfrac{3}{4}.[/tex]
Thus, the required equation of the line in slope-intercept form is [tex]y=\dfrac{1}{2}x+\dfrac{3}{4}.[/tex]
