The lines $y = m_{1} x, y = m_{2} x$ and $y = m_{3} x$ make equal intercepts on the line $x + y = 1,$ then

2 (1 + m_{1}) (1 + m_{3}) = (1 + m_{2}) (2 + m_{1} + m_{3})

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(1 + m_{1}) (1 + m_{3}) = (1 + m_{2}) (1 + m_{1} + m_{3})

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(1 + m_{1}) (1 + m_{2}) = (1 + m_{3}) (2 + m_{1} + m_{3})

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2 (1 + m_{1}) (1 + m_{3}) = (1 + m_{2}) (1 + m_{1} + m_{3})

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Solution

The correct option is B $2 (1 + m_{1}) (1 + m_{3}) = (1 + m_{2}) (2 + m_{1} + m_{3})$
Substituting $y = m x$ in $x + y = 1$ we get $x + m x = 1 \to x = \frac{1}{1 + m} \to y = \frac{m}{1 + m}$
So Co-ordinates of A,B,C are $(\frac{1}{1 + m_{1}}, \frac{m_{1}}{1 + m_{1}}) (\frac{1}{1 + m_{2}}, \frac{m_{2}}{1 + m_{2}}) (\frac{1}{1 + m_{3}}, \frac{m_{3}}{1 + m_{3}})$ respectively
Since $A B = B C$
B is midpoint of AC and lies on $y = m_{2} x$
$∴ \frac{1}{2} \times (\frac{m_{1}}{1 + m_{1}} + \frac{m_{3}}{1 + m_{3}}) = m_{2} \times \frac{1}{2} \times (\frac{1}{1 + m_{1}} + \frac{1}{1 + m_{3}})$
On solving, we get
$2 (1 + m_{1}) (1 + m_{3}) = (1 + m_{2}) (2 + m_{1} + m_{3})$

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The lines y=m1x,y=m2x and y=m3x make equal intercepts on the line x+y=1, then

The lines $y = m_{1} x, y = m_{2} x$ and $y = m_{3} x$ make equal intercepts on the line $x + y = 1,$ then