Question

If $A(1,p^2),B(0,1)$ 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. $\frac{1}{\sqrt{3}}$
• B. $\frac{-1}{\sqrt{3}}$
• C. $\frac{1}{\sqrt{2}}$
• D. None

Answer 1

The correct option is A $\frac{1}{\sqrt{3}}$

$Area=\frac{1}{2}\left|\begin{array}{ccc}1& p^2& 1\\ 0& 1& 1\\ p& 0& 1\end{array}\right|$

$=\frac{1}{2}\left(1\right(1-0)-0+p(p^2-1\left)\right)$

$=\frac{1}{2}(p^3-p+1)$

$\frac{d\left(Area\right)}{dp}=\frac{1}{2}(3p^2-1)=0$

$\frac{d^{2}Area}{d{p}^2}=\frac{1}{2}\left(6p\right)>0$ when $p=\frac{1}{\sqrt{3}}$

Area =min at $p=\frac{1}{\sqrt{3}}$