Answer
$y=-\dfrac{1}{3}x+\dfrac{8}{3}$
Work Step by Step
Using $y-y_1=\dfrac{y_1-y_2}{x_1-x_2}(x-x_1)$ or the Two-Point Form of linear equations, where
\begin{array}{l}\require{cancel}
x_1=
5
,\\x_2=
8
,\\y_1=
1
,\\y_2=
0
,\end{array}
the equation of the line is
\begin{array}{l}\require{cancel}
y-1=\dfrac{1-0}{5-8}(x-5)
\\\\
y-1=\dfrac{1}{-3}(x-5)
\\\\
y-1=-\dfrac{1}{3}(x-5)
.\end{array}
In $y=mx+b$ or the Slope-Intercept Form, the equation above is equivalent to
\begin{array}{l}\require{cancel}
y-1=-\dfrac{1}{3}(x-5)
\\\\
y-1=-\dfrac{1}{3}(x)-\dfrac{1}{3}(-5)
\\\\
y-1=-\dfrac{1}{3}x+\dfrac{5}{3}
\\\\
y=-\dfrac{1}{3}x+\dfrac{5}{3}+1
\\\\
y=-\dfrac{1}{3}x+\dfrac{5}{3}+\dfrac{3}{3}
\\\\
y=-\dfrac{1}{3}x+\dfrac{8}{3}
.\end{array}
