Question

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 $ABC$ is the minimum is

• A. $1/\sqrt{3}$
• B. $-1/\sqrt{3}$
• C. $1/\sqrt{2}$
• D. none of these

Answer 1

Answer: (D)

$A=\frac{1}{2}\left|\begin{array}{ccc}1& p^2& 1\\ 0& 1& 1\\ p& 0& 1\end{array}\right|$
$=\frac{1}{2}\left[1(1-0)+p\left(p^2-1\right)\right]$
$=\frac{1}{2}\left(p^3-p+1\right)$
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 $(a),(b),(c)$.