Answer
$\text{a) Slope-Intercept Form: }
y=-x+2
\\\\\text{b) Standard Form: }
x+y=2$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To find the equation of the line that passes through the points $(
0,2
)$ and $(
2,0
),$ (see Exercise 18) use the Two-Point Form of linear equations. Give the equation in the Slope-Intercept Form and in the Standard Form.
$\bf{\text{Solution Details:}}$
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=0
,\\x_2=2
,\\y_1=2
,\\y_2=0
,\end{array}
the equation of the line is
\begin{array}{l}\require{cancel}
y-y_1=\dfrac{y_1-y_2}{x_1-x_2}(x-x_1)
\\\\
y-2=\dfrac{2-0}{0-2}(x-0)
\\\\
y-2=\dfrac{2}{-2}(x)
\\\\
y-2=-1(x)
\\\\
y-2=-x
.\end{array}
Using the properties of equality, in the form $y=mx+b$ or the Slope-Intercept Form, the equation above is equivalent to
\begin{array}{l}\require{cancel}
y-2=-x
\\\\
y=-x+2
.\end{array}
Using the properties of equality, in the form $ax+by=c$ or the Standard Form, the equation above is equivalent to
\begin{array}{l}\require{cancel}
y=-x+2
\\\\
x+y=2
.\end{array}
The equation of the line is
\begin{array}{l}\require{cancel}
\text{a) Slope-Intercept Form: }
y=-x+2
\\\\\text{b) Standard Form: }
x+y=2
.\end{array}