Answer
The equation of the required line is: $y=\frac{1}{3}x+\frac{10}{3}$.
Work Step by Step
An equation of line that passes through the points $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ is given by,
$\left| \begin{matrix}
x & y & 1 \\
{{x}_{1}} & {{y}_{1}} & 1 \\
{{x}_{2}} & {{y}_{2}} & 1 \\
\end{matrix} \right|=0$
Therefore, the equation of the line that passes through the points $\left( -1,3 \right)$ and $\left( 2,4 \right)$ is given by,
$\left| \begin{matrix}
x & y & 1 \\
-1 & 3 & 1 \\
2 & 4 & 1 \\
\end{matrix} \right|=0$
Expand along the first row to get,
$x\left| \begin{matrix}
3 & 1 \\
4 & 1 \\
\end{matrix} \right|-y\left| \begin{matrix}
-1 & 1 \\
2 & 1 \\
\end{matrix} \right|+1\left| \begin{matrix}
-1 & 3 \\
2 & 4 \\
\end{matrix} \right|=0$
Simplify the above expression as below:
$\begin{align}
& x\left( 3-4 \right)-y\left( -1-2 \right)+1\left( -4-6 \right)=0 \\
& -x+3y-10=0 \\
& -x+3y=10
\end{align}$
This can be written as: $y=\frac{1}{3}x+\frac{10}{3}$
Thus, the equation of the required line is: $y=\frac{1}{3}x+\frac{10}{3}$
