Answer
The equation of the required line is: $y=-\frac{11}{5}x+\frac{8}{5}$.
Work Step by Step
An equation of the 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 $\left( 3,-5 \right)$ and $\left( -2,6 \right)$ is given by,
$\left| \begin{matrix}
x & y & 1 \\
3 & -5 & 1 \\
-2 & 6 & 1 \\
\end{matrix} \right|=0$
Expand along the first row to get,
$x\left| \begin{matrix}
-5 & 1 \\
6 & 1 \\
\end{matrix} \right|-y\left| \begin{matrix}
3 & 1 \\
-2 & 1 \\
\end{matrix} \right|+1\left| \begin{matrix}
3 & -5 \\
-2 & 6 \\
\end{matrix} \right|=0$
Simplify the above expression as below:
$\begin{align}
& x\left( -5-6 \right)-y\left( 3+2 \right)+1\left( 18-10 \right)=0 \\
& -11x-5y+8=0 \\
& 11x+5y-8=0
\end{align}$
This can be written as: $5y=-11x+8$ or $y=-\frac{11}{5}x+\frac{8}{5}$.
Thus, the equation of the required line is: $y=-\frac{11}{5}x+\frac{8}{5}$
