Question

The three lines $lx + my + n = 0$, $mx + ny + l = 0$, $nx + ly + m = 0$ are concurrent if
A). $l = m + n$
B). $m = l + n$
C). $n = l + m$
D). $l + m + n = 0$

Answer 1

In order to find the condition for the three lines given to be concurrent, we need to know what exactly concurrency is. Concurrent lines are lines that are intersecting at a point. For three lines the condition satisfies that the determinant of the coefficient of the lines should be equal to zero.

Complete step-by-step solution:
We are given three lines that are as follows:
$lx + my + n = 0$
$mx + ny + l = 0$
$nx + ly + m = 0$
And, also these three lines are concurrent that implies the three lines given intersect at a single point.
Now, considering three straight lines whose equations are as follows:
${a_1}x + {b_1}y + {c_1} = 0$
${a_2}x + {b_2}y + {c_2} = 0$
${a_3}x + {b_3}y + {c_3} = 0$
According to the conditions, these three lines are said to be concurrent if the determinant of the coefficient of the three lines forming a 3x3 matrix gives zero as a result.
The determinant of the coefficient numerically written as:
\left| {\begin{array}{*{20}{c}} {{a_1}}&{{b_1}}&{{c_1}} \\\ {{a_2}}&{{b_2}}&{{c_2}} \\\ {{a_3}}&{{b_3}}&{{c_3}} \end{array}} \right| = 0
Similarly, the three lines we are given are already concurrent that means their determinant is equal to zero:
So, equating and writing the coefficient of the lines given in the determinant form as:
\left| {\begin{array}{*{20}{c}} l&m;&n; \\\ m&n;&l; \\\ n&l;&m; \end{array}} \right| = 0
Solving the determinant using Row 1, and we get:
$\Rightarrow l\left( {nm - {l^2}} \right) - m\left( {{m^2} - nl} \right) + n\left( {ml - {n^2}} \right) = 0$
Opening the parenthesis, and we get:
$\Rightarrow lmn - {l^3} - {m^3} + lmn + lmn - {n^3} = 0$
On further solving, we get:
$\Rightarrow 3lmn - {l^3} - {m^3} - {n^3} = 0$
Subtracting both sides by $3lmn$, we get:
$\Rightarrow 3lmn - {l^3} - {m^3} - {n^3} - 3lmn = - 3lmn$
$\Rightarrow - {l^3} - {m^3} - {n^3} = - 3lmn$
Taking $- 1$ common from the above equation and we get:
$\Rightarrow - \left( {{l^3} + {m^3} + {n^3}} \right) = - 3lmn$
Cancelling out the negative one:
$\Rightarrow {l^3} + {m^3} + {n^3} = 3lmn$ ……(1)
Now, from the formula ${{\mathbf{a}}^{\mathbf{3}}}\; + \;{{\mathbf{b}}^{\mathbf{3}}}\; + \;{{\mathbf{c}}^{\mathbf{3}}}\; - \;{\mathbf{3abc}}\; = {\text{ }}(a{\text{ }} + \;b + \;c)({a^2}\; + \;{b^2}\; + \;{c^2}\; - {\text{ }}ab{\text{ }} - {\text{ }}bc{\text{ }} - {\text{ }}ca)$, the condition says, if ${\text{ }}(a{\text{ }} + \;b + \;c)$ is equal to zero, that gives ${{\mathbf{a}}^{\mathbf{3}}}\; + \;{{\mathbf{b}}^{\mathbf{3}}}\; + \;{{\mathbf{c}}^{\mathbf{3}}}\; = 3abc$.
From equation (1), we can see that we already have ${l^3} + {m^3} + {n^3} = 3lmn$ that implies ${\text{ }}(l + \;m + \;n)$ should be equal to zero, numerically written as:
${\text{ }}(l + \;m + \;n) = 0{\text{ }}$
in case of $\left( {{\text{l}} \ne {\text{m}} \ne {\text{n}}} \right)$.
Therefore, the three lines $lx + my + n = 0$, $mx + ny + l = 0$, $nx + ly + m = 0$ are concurrent if ${\text{ }}(l + \;m + \;n) = 0{\text{ }}$.
Hence, Option (D) is correct.

Note: Remember parallel lines are never concurrent because according to the definition of concurrency, lines are concurrent if they intersects through one point but for parallel lines they do not meet at any single point, Hence, only intersecting and non-parallel lines are said to be concurrent.