JEE MainMathematicsEasyProperties of triangle

Given A (0,0) and B(x,y) with x∈(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 slope
of BC equals m_2 where 0 < m_2< m_1. If the area of the triangle
ABC can be expressed as (m_1 - m_2) f(x), then the largest possible
value of f (x) is:
fd A (0,0) rFkk B(x,y) gS tcfd x∈(0,1 ) rFkk y > 0 ekuk
çdkj gS fd BC dh ço.krk m_2 ds cjkcj tgk¡ 0 < m_2< m_1 ABC
=Qy (m_1 - m_2) f(x),

Show answer and solution

A(0,0)B(x,y)C\left( 1,y_{1} \right)

m_{1} = \frac{y}{x},m_{2} = \frac{y_{1} - y}{1 - x} \Rightarrow y_{1} = m_{2} - m_{2}x + y

\Delta = \frac{1}{2}\left| y - {xy}_{1} \right| = \frac{1}{2}\left| m_{1}x - m_{2}x + m_{2}x^{2} - m_{1}x^{2} \right| = \left( m_{1} - m_{2} \right)\frac{\left| x^{2} - x \right|}{2}

f(x) = \frac{\left| x^{2} - x \right|}{2}

f(x)max = \frac{1}{8}

Source: Grass-G / M-26 Mathematics Paper 1 15 Jan.docx

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