the lines Lx+my+n=0, mx+ny+L=0 and nx+Ly+m=0 are concurrent if
1) L+m+n=0
2) L+m-n=0
3) L-m+n=0
4)L^2+m^2+n^2=Lm+mn+nL

Dear Student,

Please find below the solution to the asked query :

) If the three lines a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0 and a₃x + b₃y + c₃ = 0 are concurrent at a point, then the determinant given by:
|a₁ b₁ c₁|
|a₂ b₂ c₂| = 0
|a₃ b₃ c₃|

ii) Applying this to the given set of 3 lines, the determinant

|l m n|
|m n l|
|n l m|

has to be zero for them to be concurrent.

iii) Applying properties of determinant, the above reduces to:
|1 1 1|*(l + m + n)
|m n l|
|n l m|

On evaluation this = -(l + m + n)(m² + n² + l² - mn - nl - ln)

= -(1/2)(l + m + n)(2m² + 2n² + 2l² - 2mn - 2nl - 2ln)

= -(1/2)(l + m + n){(m - n)² + (n - l)² + (l - m)²}

Since, l, m, n are different, (m - n), (n - l) and (l - m) not equal to zero.

Hence for the determinant to be zero, only (l + m + n) has to be zero.

Thus if (l + m + n) = 0, the 3 lines are concurrent at a point.


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