The area of the triangle with vertices (1, 4), (2, 7) and (4, 13) is :

The correct answer is Option (1) → 0

area = $\frac{1}{2}\begin{vmatrix}1&4&1\\2&7&1\\4&13&1\end{vmatrix}$

$R_3→R_3-R_2,R_2→R_2-R_1$

$\frac{1}{2}\begin{vmatrix}1&4&1\\1&3&0\\2&6&0\end{vmatrix}=\frac{6-6}{2}=0$