Using determinants, find the area of the triangle whose vertices

$\Delta=\frac{1}{2}\left|\begin{array}{ccc}1 & 4 & 1 \\ 2 & 3 & 1 \\ -5 & -3 & 1\end{array}\right|$

$=\frac{1}{2}\left|\begin{array}{ccc}1 & 4 & 1 \\ 1 & -1 & 0 \\ -5 & -3 & 1\end{array}\right| \quad$ [Applying $R_{2} \rightarrow R_{2}-R_{1}$ ]

$=\frac{1}{2}\left|\begin{array}{ccc}1 & 4 & 1 \\ 1 & -1 & 0 \\ -6 & -7 & 0\end{array}\right| \quad$ [Applying $R_{3} \rightarrow R_{3}-R_{1}$ ]

$=\frac{1}{2}\left|\begin{array}{cc}1 & -1 \\ -6 & -7\end{array}\right|$

$=\frac{1}{2}(-7-6)$

$=\frac{13}{2}$ square units $\quad[\because$ Area cannot be negative $]$

Therefore, $(1,4),(2,3)$ and $(-5,-3)$ are not collinear because, $\left|\begin{array}{ccc}1 & 4 & 1 \\ 2 & 3 & 1 \\ -5 & -3 & 1\end{array}\right|$ is not equal to 0 .