Question

Using determinants, find the area of the triangle whose vertices are (1, 4), (2, 3) and (−5, −3). Are the given points collinear?

Answer 1

\[∆ = \frac{1}{2}\begin{vmatrix}1 & 4 & 1 \\ 2 & 3 & 1 \\ - 5 & - 3 & 1\end{vmatrix}\] 
\[ = \frac{1}{2}\begin{vmatrix}1 & 4 & 1 \\ 1 & - 1 & 0 \\ - 5 & - 3 & 1\end{vmatrix} \left[\text{ Applying }R_2 \to R_2 - R_1 \right]\]
\[ = \frac{1}{2}\begin{vmatrix}1 & 4 & 1 \\ 1 & - 1 & 0 \\ - 6 & - 7 & 0\end{vmatrix} \left[\text{ Applying }R_3 \to R_3 - R_1 \right]\] 
\[ = \frac{1}{2}\begin{vmatrix}1 & - 1 \\ - 6 & - 7\end{vmatrix}\]
\[ = \frac{1}{2}\left( - 7 - 6 \right)\] 
\[ = \frac{13}{2}\text{ square units }\left[ \because\text{ Area cannot be negative }\right]\]
Therefore, (1, 4), (2, 3) and (−5, −3) are not collinear because,

\[\begin{vmatrix}1 & 4 & 1 \\ 2 & 3 & 1 \\ - 5 & - 3 & 1\end{vmatrix}\]  is not equal to 0.