#### Answer

$\color{blue}{y=-7x+15}$.

#### Work Step by Step

Recall:
(1) The slope $m$ of a line passing through points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the formula
$$m=\dfrac{y_2-y_1}{x_2-x_1}$$
(2) The slope-intercept form of a line's equation is $y=mx+b$ where $m$=slope and $(0, b)$ is the $y$-intercept.
Solve for the slope of the line using the formula above and the points $(2, 1)$ and $(4, -13)$:
\begin{align*}
m&=\frac{y_2-y_1}{x_2-x_1}\\
m&=\frac{-13-1}{4-2}\\\\
m&=\frac{-14}{2}\\\\
m&=-7
\end{align*}
Hence, the tentative equation fo the line that contains the given points is:
$$y=-7x+b$$
Solve for the $b$ by substituting the $x$ and $y$ values of the point $(2, 1)$ into the tentative equation above to obtain:
\begin{align*}
y&=-7x+b\\
1&=-7(2)+b\\
1&=-14+b\\
1+14&=b\\
15&=b\\
\end{align*}
Therefore, the equation of the line that contains the given points in the table is $\color{blue}{y=-7x+15}$.