#### Answer

$x-4y=-13$

#### 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 given characteristcs:
\begin{array}{l}\require{cancel}
\text{through }
(-1,3)
\text{ and }
(3,4)
.\end{array}
Use the properties of equality to express the equation 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=-1
,\\x_2=3
,\\y_1=3
,\\y_2=4
,\end{array}
the equation of the line is
\begin{array}{l}\require{cancel}
y-3=\dfrac{3-4}{-1-3}(x-(-1))
\\\\
y-3=\dfrac{-1}{-4}(x+1)
\\\\
y-3=\dfrac{1}{4}(x+1)
.\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}
4(y-3)=\left[\dfrac{1}{4}(x+1)\right]4
\\\\
4(y-3)=1(x+1)
\\\\
4(y)+4(-3)=1(x)+1(1)
\\\\
4y-12=x+1
\\\\
-x+4y=1+12
\\\\
-x+4y=13
\\\\
-1(-x+4y)=-1(13)
\\\\
x-4y=-13
.\end{array}