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 the triangle $ABC$ is minimum is

• A. $\frac{1}{\sqrt{3}}$
• B. $-\frac{1}{\sqrt{3}}$
• C. $\frac{1}{\sqrt{2}}$ or $-\frac{1}{\sqrt{3}}$
• D. $none$

Answer 1

The correct option is D $none$

Equation of line $BC$ is : $\frac{x}{p}+\frac{y}{1}=1$

intercept form of a straight line is $\frac{x}{a}+\frac{y}{b}=1$, here $a$ and $b$ are respectively intercepts on $x$-axis and $y$-axis.

$BC=\sqrt{1+p^2}$.

Perpendicular distance of $A$ from $BC$ is$:\left|\frac{\frac{1}{p}+p^2-1}{\sqrt{(\frac{1}{p}{)}^2+1}}\right|=\frac{|1+p^3-p|}{\sqrt{p^2+1}}$

Area $=\frac{1}{2}\times$ base $\times$ height $=\frac{1}{2}\times \sqrt{1+p^2}\times \frac{|1+p^3-p|}{\sqrt{p^2+1}}=\frac{1}{2}\times |1+p^3-p|$

$|1+p^3-p|$ can have value $0$, so minimum is $0$ and (a),(b),(c) dont satisfy the condition.