Question

Given $A(0,0)$ and $B(x,y)$ with $x\in (0,1)$ and $y>0$. Let the slope of the line $AB$ equals $m_1$. Point $C$ lies on the line $x=1$ such that the slope of $BC$ equals $m_2$ where $0<m_2<m_1$. If the area of the triangle $ABC$ 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 $△ABC$
$=\frac{1}{2}\left|\begin{array}{ccc}0& 0& 1\\ x& m_{1}x& 1\\ 1& c& 1\end{array}\right|$
$=\frac{1}{2}\left[cx-m_{1}x\right]$
$=\frac{1}{2}\left|\left[\left(\left(m_1-m_2\right)x+m_2\right)x-m_{1}x\right]\right|$
$=\frac{1}{2}\left|\left[\left(m_1-m_2\right)x^2+m_{2}x-m_{1}x\right]\right|$
$=\frac{1}{2}\left(m_1-m_2\right)\left(x-x^2\right)(x>x^2$ in $(0,1))$
Hence, $f(x)=\frac{1}{2}\left(x-x^2\right)$
$\therefore f(x){]}_{\max }=\frac{1}{8}$ when $x=\frac{1}{2}$