## Precalculus (6th Edition) Blitzer

The equation of the required line is: $y=\frac{1}{3}x+\frac{10}{3}$.
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: $y=\frac{1}{3}x+\frac{10}{3}$