The three lines lx + my + n = 0, mx + ny + l = 0, nx + ly + m = 0 are concurrent if

1) l = m + n

2) m = l + n

3) n = l + m

4) l + m + n = 0

Solution:

Given lx + my + n = 0, mx + ny + l = 0, nx + ly + m = 0 are concurrent.

So

\(\begin{array}{l}\begin{vmatrix} a_{1} & b_{1}&c_{1}\ a_{2}& b_{2} & c_{2}\ a_{3}& b_{3} & c_{3} \end{vmatrix} = 0\end{array} \)

=>

\(\begin{array}{l}\begin{vmatrix} l & m&n\ m& n & l\ n& l & m \end{vmatrix} = 0\end{array} \)

=> l(mn – l2) -m(m2 – nl) +n(ml – n2) = 0

=> 3lmn – (l3 + m3 + n3) = 0

=> l3 + m3 + n3– 3lmn = 0

=> l + m + n = 0

Hence option (4) is the answer.

Was this answer helpful?

 
   

4 (8)

(5)
(9)

Choose An Option That Best Describes Your Problem

Thank you. Your Feedback will Help us Serve you better.

Leave a Comment

Your Mobile number and Email id will not be published. Required fields are marked *

*

*

Ask
Question