#### Answer

$x+2y=6$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
Use the Two-Point Form of linear equations to find the equation of the line with the following characteristics:
\begin{array}{l}\require{cancel}
\text{Through }
(0,3) \text{ and } (-2,4)
.\end{array}
Express the answer in 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=3
,\\y_2=4
,\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-3=\dfrac{3-4}{0-(-2)}(x-0)
\\\\
y-3=\dfrac{3-4}{0+2}(x)
\\\\
y-3=\dfrac{-1}{2}(x)
\\\\
y-3=-\dfrac{1}{2}x
.\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-3=-\dfrac{1}{2}x
\\\\
2(y-3)=2\left( -\dfrac{1}{2}x \right)
\\\\
2y-6=-x
\\\\
x+2y=6
.\end{array}