Question

The three lines $\mathrm{lx}+\mathrm{my}+\mathrm{n}=0,\mathrm{mx}+\mathrm{ny}+\mathrm{l}=0,\mathrm{nx}+\mathrm{ly}+\mathrm{m}=0$ are concurrent if

• A. $\mathrm{l}=\mathrm{m}+\mathrm{n}$
• B. $\mathrm{m}=\mathrm{l}+\mathrm{n}$
• C. $\mathrm{n}=\mathrm{l}+\mathrm{m}$
• D. $\mathrm{l}+\mathrm{m}+\mathrm{n}=0$

Answer 1

The correct option is D

$\mathrm{l}+\mathrm{m}+\mathrm{n}=0$

Condition for concurrent of three lines :

Three lines $\mathrm{lx}+\mathrm{my}+\mathrm{n}=0,\mathrm{mx}+\mathrm{ny}+\mathrm{l}=0,\mathrm{nx}+\mathrm{ly}+\mathrm{m}=0$ are concurrent.

So, that means those three lines meet at a common point and $\left|\begin{array}{ccc}\mathrm{l}& \mathrm{m}& \mathrm{n}\\ \mathrm{m}& \mathrm{n}& \mathrm{l}\\ \mathrm{n}& \mathrm{l}& \mathrm{m}\end{array}\right|=0$

Thus,

$$\begin{gathered} \mathrm{l}(\mathrm{mn}-{\mathrm{l}}^2)-\mathrm{m}({\mathrm{m}}^2-\mathrm{nl})+\mathrm{n}(\mathrm{ml}-{\mathrm{n}}^2)=0 \\ \Rightarrow 3\mathrm{lmn}-({\mathrm{l}}^3+{\mathrm{m}}^3+{\mathrm{n}}^3)=0 \\ \Rightarrow {\mathrm{l}}^3+{\mathrm{m}}^3+{\mathrm{n}}^3-3\mathrm{lmn}=0 \\ \Rightarrow \mathrm{l}+\mathrm{m}+\mathrm{n}=0[{\mathrm{l}}^3+{\mathrm{m}}^3+{\mathrm{n}}^3=3\mathrm{lmn}\Rightarrow \mathrm{l}+\mathrm{m}+\mathrm{n}=0] \end{gathered}$$

Hence option (D) is the correct option.