Question

Given $A(0,0)$ and $B(x, y)$ with $x \in(0,1)$ and $y>0$. Let the slope of the line $A B$ equals $m_{1}$. Point $C$ lies on the line $x=1$ such that the slope of $B C$ equals $m_{2}$ where $0 < m_{2} < m_{1}$. If the area of the triangle $A B C$ can be expressed as $\left(m_{1}-m_{2}\right) f(x)$, then the largest possible value of $f(x)$ is

• A. $1$
• B. $1 / 2$
• C. $1 / 4$
• D. $1 / 8$

Answer 1

Answer: (D)

Let the coordinates of $C$ be $(1, c)$.
$m_{2} =\frac{c-y}{1-x} $
or $ m_{2} =\frac{c-m_{1} x}{1-x}\left(\text { as } y=m_{1} x\right)$
or $ m_{2}-m_{2} x=c-m_{1} x$
or $\left(m_{1}-m_{2}\right) x=c-m_{2}$
$\therefore c=\left(m_{1}-m_{2}\right) x+m_{2}$ ..... (i)
Now, area of $\triangle A B C$
$=\frac{1}{2}\begin{vmatrix}0 & 0 & 1 \\ x & m_{1} x & 1 \\ 1 & c & 1\end{vmatrix}$
$=\frac{1}{2}\left[c x-m_{1} x\right]$
$=\frac{1}{2}\left|\left[\left(\left(m_{1}-m_{2}\right) x+m_{2}\right) x-m_{1} x\right]\right|$
$=\frac{1}{2}\left|\left[\left(m_{1}-m_{2}\right) x^{2}+m_{2} x-m_{1} x\right]\right|$
$=\frac{1}{2}\left(m_{1}-m_{2}\right)\left(x-x^{2}\right) \,\,\,\left(x>x^{2}\right.$ in $\left.(0,1)\right)$
Hence, $f(x)=\frac{1}{2}\left(x-x^{2}\right)$
$\therefore f(x)]_{\max }=\frac{1}{8}$ when $x=\frac{1}{2}$