#### Answer

$\text{slope-intercept form: }
y=\dfrac{1}{4}x-\dfrac{15}{4}
\\\\
\text{standard form: }
x-4y=15$

#### Work Step by Step

Using $y-y_1=m(x-x_1)$ or the point-slope form of linear equations, the equation of the line passing through $(
7,-2
)$ and with a slope of $m=
\dfrac{1}{4}
$ is
\begin{array}{l}\require{cancel}
y-(-2)=\dfrac{1}{4}(x-7)
\\\\
y+2=\dfrac{1}{4}(x-7)
.\end{array}
In the form $y=mx+b$, the equation above is equivalent to
\begin{array}{l}\require{cancel}
y+2=\dfrac{1}{4}(x-7)
\\\\
y+2=\dfrac{1}{4}x-\dfrac{7}{4}
\\\\
y=\dfrac{1}{4}x-\dfrac{7}{4}-2
\\\\
y=\dfrac{1}{4}x-\dfrac{7}{4}-\dfrac{8}{4}
\\\\
y=\dfrac{1}{4}x-\dfrac{15}{4}
.\end{array}
In the form $Ax+By=C$, the equation above is equivalent to
\begin{array}{l}\require{cancel}
y=\dfrac{1}{4}x-\dfrac{15}{4}
\\\\
4(y)=\left(\dfrac{1}{4}x-\dfrac{15}{4}\right)4
\\\\
4y=x-15
\\\\
-x+4y=-15
\\\\
x-4y=15
.\end{array}
Hence, the different forms of the equation of the line with the given conditions are
\begin{array}{l}\require{cancel}
\text{slope-intercept form: }
y=\dfrac{1}{4}x-\dfrac{15}{4}
\\\\
\text{standard form: }
x-4y=15
.\end{array}