Area of the triangle ABC with vertices $\left(a, a^2\right),\left(b, b^2\right)$ and $\left(c, c^2\right)$ is equal to:

$\frac{1}{2}|(a-b)(b-c)(c-a)|$

The correct answer is Option (3) → $\frac{1}{2}|(a-b)(b-c)(c-a)|$

area = $\frac{1}{2}\begin{vmatrix}a&a^2&1\\b&b^2&1\\c&c^2&1\end{vmatrix}$

$R_1→R_1-R_2,R_2→R_2-R_3$

$=\frac{1}{2}\begin{vmatrix}a-b&a^2-b^2&0\\b-c&b^2-c^2&0\\c&c^2&1\end{vmatrix}$

$=\frac{(a-b)(b-c)}{2}\begin{vmatrix}1&a+b&0\\1&b+c&0\\c&c^2&1\end{vmatrix}$

$=\frac{(a-b)(b-c)}{2}(b+c-a-b)$

$=\frac{(a-b)(b-c)(c-a)}{2}$