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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Aicharya · Accepted Answer

Since, the point of intersecting of given lines $L_1: y=m_a x$, $L_2: y=m_b x$ and $L_3: y=m_c x$ with line $x+y=1$ are $\begin{aligned} & A\left(\frac{1}{1+m_a}, \frac{m_a}{1+m_a}ight), B\left(\frac{1}{1+m_b}, \frac{m_b}{1+m_b}ight) 	ext { and } \ & C\left(\frac{1}{1+m_c}, \frac{m_c}{1+m_c}ight) 	ext { respectively. } \end{aligned}$ It is given that, $A B=B C \Rightarrow A B^2=B C^2$ $\begin{aligned} &\Rightarrow\left(\frac{1}{1+m_a}-\frac{1}{1+m_b}ight)^2+\left(\frac{m_a}{1+m_a}-\frac{m_b}{1+m_b}ight)^2 \ &=\left(\frac{1}{1+m_b}-\frac{1}{1+m_c}ight)^2+\left(\frac{m_b}{1+m_b}-\frac{m_c}{1+m_c}ight)^2 \ &\Rightarrow \quad \frac{\left(m_b-m_aight)^2}{\left(1+m_aight)^2\left(1+m_bight)^2}+\frac{\left(m_a-m_bight)^2}{\left(1+m_aight)^2\left(1+m_bight)^2} \ &=\frac{\left(m_c-m_bight)^2}{\left(1+m_bight)^2\left(1+m_cight)^2}+\frac{\left(m_b-m_cight)^2}{\left(1+m_bight)^2\left(1+m_cight)^2} \ &\Rightarrow \quad 2 \frac{\left(m_a-m_bight)^2}{\left(1+m_aight)^2\left(1+m_bight)^2}=2 \frac{\left(m_b-m_cight)^2}{\left(1+m_bight)^2\left(1+m_cight)^2} \ &\Rightarrow \quad\left(m_a-m_bight)\left(1+m_cight)=\left(m_b-m_cight)\left(1+m_aight) \ &\Rightarrow \quad\left[\left(1+m_aight)-\left(1+m_bight)ight]\left(1+m_cight) =\left[\left(1+m_bight)-\left(1+m_cight)ight]\left(1+m_aight) \ &\Rightarrow \quad 2\left(1+m_aight)\left(1+m_cight)=\left(1+m_bight)\left(1+m_c+1+m_aight) \ &\Rightarrow \quad 2\left(1+m_aight)\left(1+m_cight)=\left(1+m_bight)\left(2+m_a+m_cight) \end{aligned}$

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…

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