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Q.
If $A\left(1, p^{2}\right), B(0,1)$, and $C(p, 0)$ are the coordinates of three points, then the value of $p$ for which the area of triangle $A B C$ is the minimum is
Straight Lines
Solution:
$A=\frac{1}{2} \begin{vmatrix}1 & p^{2} & 1 \\0 & 1 & 1 \\p & 0 & 1\end{vmatrix}$
$= \frac{1}{2}\left[1(1-0)+p\left(p^{2}-1\right)\right] $
$=\frac{1}{2}\left(p^{3}-p+1\right)$
Hence, $A=\frac{1}{2}\left|p^{3}-p+1\right|$.
Now, the minimum value of modulus is zero.
Since $A(p)$ is cubic, it must vanish for some $p$ other than given in $(a),(b),(c)$.