#### Answer

$y=\dfrac{1}{2}x+1$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
Use the Two-Point Form of linear equations to find the equation of the line passing through $(
6,4)$ and $(-4,-1).$
$\bf{\text{Solution Details:}}$
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=6
,\\x_2=-4
,\\y_1=4
,\\y_2=-1
,\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-4=\dfrac{4-(-1)}{6-(-4)}(x-6)
\\\\
y-4=\dfrac{4+1}{6+4}(x-6)
\\\\
y-4=\dfrac{5}{10}(x-6)
\\\\
y-4=\dfrac{1}{2}(x-6)
\\\\
y-4=\dfrac{1}{2}(x)+\dfrac{1}{2}(-6)
\\\\
y-4=\dfrac{1}{2}x-3
\\\\
y=\dfrac{1}{2}x-3+4
\\\\
y=\dfrac{1}{2}x+1
.\end{array}