Straight Line Question 279

Question: If the transversal y = mr x; r = 1, 2, 3 cut off equal intercepts on the transversal $ x+y=1, $ then $ 1+m_1, $ $ 1+m_2, $ $ 1+m_3 $ are in

Options:

A)A. P.

B)G. P.

C)H. P.

D)None of these

Show Answer

Answer:

Correct Answer: C

Solution:

  • 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 $ ( \frac{1}{1+m_1},\frac{m_1}{1+m_1} ), $ $ ( \frac{1}{1+m_2},\frac{m_2}{1+m_2} ) $ and $ ( \frac{1}{1+m_3},\frac{m_3}{1+m_3} ) $ By hypothesis, $ {{( \frac{1}{1+m_1}-\frac{1}{1+m_2} )}^{2}}+{{( \frac{m_1}{1+m_1}-\frac{m_2}{1+m_2} )}^{2}} $ = $ {{( \frac{1}{1+m_2}-\frac{1}{1+m_3} )}^{2}}+{{( \frac{m_2}{1+m_2}-\frac{m_3}{1+m_2} )}^{2}} $
    Þ $ \frac{m_2-m_1}{1+m_1}=\frac{m_3-m_2}{1+m_3} $ or $ \frac{1+m_2}{1+m_1}-1=1-\frac{1+m_2}{1+m_3} $
    Þ $ \frac{1+m_2}{1+m_1}+\frac{1+m_2}{1+m_3}=2 $ Þ $ 1+m_2=\frac{2(1+m_1)(1+m_3)}{(1+m_1)+(1+m_3)} $
    Þ $ 1+m_1,1+m_2,1+m_3 $ are in H.P.