## Precalculus (6th Edition) Blitzer

The equation of the required line is: $y=-\frac{11}{5}x+\frac{8}{5}$.
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: $y=-\frac{11}{5}x+\frac{8}{5}$