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:
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Solution
Sol. [Key Concept: Express area $\bigtriangleup ABC$ as a function of the base position $x$ and maximise $x - x^{2}$ ⇒ value $\frac{1}{8}$.]
Let the coordinates of $C$ be $(1,c)$
$$\begin{matrix} & & m_{2} = \frac{c - y}{1 - x};\ m_{2} = \frac{c - m_{1}x}{1 - x} \\ & & m_{2} - m_{2}x = c - m_{1}x \\ & & \left( m_{1} - m_{2} \right)x = c - m_{2} \\ & c = \left( m_{1} - m_{2} \right)x + m_{2} & \text{(1)} \end{matrix}$$
now area of $\bigtriangleup ABC = \frac{1}{2}\left| \begin{matrix} 0 & 0 & 1 \\ x & m_{1}x & 1 \\ 1 & c & 1 \end{matrix} \right|$
$$\begin{matrix} & \ = \frac{1}{2}\left\lbrack cx - m_{1}x \right\rbrack \\ & \ = \frac{1}{2}\left| \left\lbrack \left( \left( m_{1} - m_{2} \right)x + m_{2} \right)x - m_{1}x \right\rbrack \right| \\ & \ = \frac{1}{2}\left| \left\lbrack \left( m_{1} - m_{2} \right)x^{2} + m_{2}x - m_{1}x \right\rbrack \right| \\ & \ = \frac{1}{2}\left( m_{1} - m_{2} \right)\left( x - x^{2} \right) \end{matrix}$$
$$\left\lbrack \because x > x^{2}\text{~}\text{in}\text{~}(0,1\rbrack \right.\ $$
Hence, $f(x) = \frac{1}{2}\left( x - x^{2} \right)$;
$f(x)\rbrack_{\max} = \frac{1}{8}\ $ when $\ x = \frac{1}{2}$
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