If the transversals y = m_r x; r = 1, 2, 3 cu | vedclass

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(C) Solving $y = m_r x$ and $x + y = 1$,we get $x = \frac{1}{1 + m_r}$ and $y = \frac{m_r}{1 + m_r}$.Thus,the points of intersection of the three line

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$H.P.$

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(C) Solving $y = m_r x$ and $x + y = 1$,we get $x = \frac{1}{1 + m_r}$ and $y = \frac{m_r}{1 + m_r}$.Thus,the points of intersection of the three lines on the transversal are $P_r = \left( \frac{1}{1 + m_r}, \frac{m_r}{1 + m_r} \right)$ for $r = 1, 2, 3$.Since the intercepts between consecutive points are equal,the distance between $P_1$ and $P_2$ equals the distance between $P_2$ and $P_3$.Using the distance formula,the condition for equal intercepts simplifies to $\frac{1}{1 + m_1} - \frac{1}{1 + m_2} = \frac{1}{1 + m_2} - \frac{1}{1 + m_3}$ (considering the projection on the x-axis).This implies $\frac{2}{1 + m_2} = \frac{1}{1 + m_1} + \frac{1}{1 + m_3}$.Therefore,$1 + m_1, 1 + m_2, 1 + m_3$ are in $H.P.$

If the transversals $y = m_r x; r = 1, 2, 3$ cut off equal intercepts on the line $x + y = 1$,then $1 + m_1, 1 + m_2, 1 + m_3$ are in

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