## Precalculus (6th Edition) Blitzer

The points $\left( 3,-1 \right),\left( 0,-3 \right)\text{ and }\left( 12,5 \right)\text{ are collinear}\text{.}$
The points $\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)\text{ and }\left( {{x}_{3}},{{y}_{3}} \right)$ are collinear if, $\left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1 \\ {{x}_{2}} & {{y}_{2}} & 1 \\ {{x}_{3}} & {{y}_{3}} & 1 \\ \end{matrix} \right|=0$ In order to check if the points $\left( 3,-1 \right),\left( 0,-3 \right)\text{ and }\left( 12,5 \right)$ are collinear, find out the determinant below: $\left| \begin{matrix} 3 & -1 & 1 \\ 0 & -3 & 1 \\ 12 & 5 & 1 \\ \end{matrix} \right|$ Evaluate the above determinant by expanding along the first column as below: \begin{align} & \left| \begin{matrix} 3 & -1 & 1 \\ 0 & -3 & 1 \\ 12 & 5 & 1 \\ \end{matrix} \right|=3\left| \begin{matrix} -3 & 1 \\ 5 & 1 \\ \end{matrix} \right|+12\left| \begin{matrix} -1 & 1 \\ -3 & 1 \\ \end{matrix} \right| \\ & =3\left( -3-5 \right)+12\left( -1+3 \right) \\ & =3\left( -8 \right)+12\left( 2 \right) \\ & =0 \end{align} As this determinant is equal to zero, therefore, the provided points are collinear. Thus, the points $\left( 3,-1 \right),\left( 0,-3 \right)\text{ and }\left( 12,5 \right)\text{ are collinear}\text{.}$