Question

If the transversal y = m_r x; r = 1, 2, 3 cut off equal intercepts on the transversal then $1 + m _ { 1 }$ $1 + m _ { 2 }$ $1 + m _ { 3 }$ are in.

• A. A. P.
• B. G. P.
• C. H. P.
• D. None of these

Answer 1

Answer: (C)

H. P.

Answer 2

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 $\left( \frac { 1 } { 1 + m _ { 1 } } , \frac { m _ { 1 } } { 1 + m _ { 1 } } \right)$ $\left( \frac { 1 } { 1 + m _ { 2 } } , \frac { m _ { 2 } } { 1 + m _ { 2 } } \right)$ and $\left( \frac { 1 } { 1 + m _ { 3 } } , \frac { m _ { 3 } } { 1 + m _ { 3 } } \right)$

By hypothesis, $\left( \frac { 1 } { 1 + m _ { 1 } } - \frac { 1 } { 1 + m _ { 2 } } \right) ^ { 2 } + \left( \frac { m _ { 1 } } { 1 + m _ { 1 } } - \frac { m _ { 2 } } { 1 + m _ { 2 } } \right) ^ { 2 }$