Given that lines L_1: y=m_a x, L_2: y=m_b x a | vedclass

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(B) The intersection points of lines $L_1, L_2, L_3$ with $x+y=1$ are $A(\frac{1}{1+m_a}, \frac{m_a}{1+m_a})$,$B(\frac{1}{1+m_b}, \frac{m_b}{1+m_b})$,

Vedclass · Accepted Answer

$2(1+m_a)(1+m_c)=(1+m_b)(2+m_a+m_c)$

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(B) The intersection points of lines $L_1, L_2, L_3$ with $x+y=1$ are $A(\frac{1}{1+m_a}, \frac{m_a}{1+m_a})$,$B(\frac{1}{1+m_b}, \frac{m_b}{1+m_b})$,and $C(\frac{1}{1+m_c}, \frac{m_c}{1+m_c})$.Since $AB=BC$,we have $AB^2=BC^2$.Using the distance formula,$AB^2 = (\frac{1}{1+m_a} - \frac{1}{1+m_b})^2 + (\frac{m_a}{1+m_a} - \frac{m_b}{1+m_b})^2 = \frac{(m_b-m_a)^2}{(1+m_a)^2(1+m_b)^2} + \frac{(m_a-m_b)^2}{(1+m_a)^2(1+m_b)^2} = \frac{2(m_a-m_b)^2}{(1+m_a)^2(1+m_b)^2}$.Similarly,$BC^2 = \frac{2(m_b-m_c)^2}{(1+m_b)^2(1+m_c)^2}$.Equating $AB^2 = BC^2$,we get $\frac{(m_a-m_b)^2}{(1+m_a)^2} = \frac{(m_b-m_c)^2}{(1+m_c)^2}$.Taking the square root,$\frac{m_a-m_b}{1+m_a} = \pm \frac{m_b-m_c}{1+m_c}$.Taking the positive case (assuming order $A, B, C$): $\frac{m_a-m_b}{1+m_a} = \frac{m_b-m_c}{1+m_c} \Rightarrow (m_a-m_b)(1+m_c) = (m_b-m_c)(1+m_a)$.Adding $1$ to both sides of the terms: $(1+m_a - (1+m_b))(1+m_c) = (1+m_b - (1+m_c))(1+m_a)$.$(1+m_a)(1+m_c) - (1+m_b)(1+m_c) = (1+m_b)(1+m_a) - (1+m_c)(1+m_a)$.$2(1+m_a)(1+m_c) = (1+m_b)(1+m_a+1+m_c) = (1+m_b)(2+m_a+m_c)$.

Given that lines $L_1: y=m_a x, L_2: y=m_b x$ and $L_3: y=m_c x$ make equal intercepts on the line $x+y=1$,then

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