Question

Consider three equations of line:
lx + my + n = 0
mx + ny + l = 0
nx + ly + m = 0

These three lines are concurrent when?

Options are:-
[A] l + m - n = 0
[B] l + m + n = 0
[C] l - m - n = 0
[D] l - m + n = 0

Answer along with explanation.

Answer 1

$\large\underline{\sf{Solution-}}$

Given equation of lines are

$\rm \: lx + my + n = 0 - - - (1)$

$\rm \: mx + ny + l = 0 - - - (2)$

$\rm \: nx + ly + m = 0 - - - (3)$

We know, three lines

ax + by + c = 0

dx + ey + f = 0

gx + hy + i = 0 are concurrent iff

$\rm \: \: \begin{gathered}\sf \left | \begin{array}{ccc}a&b&c\\d&e&f\\g&h& i\end{array}\right | \end{gathered} \: = \: 0$

So, using this concept, we have

$\rm \: \: \begin{gathered}\sf \left | \begin{array}{ccc}l&m&n\\m&n&l\\n&l& m\end{array}\right | \end{gathered} \: = \: 0$

$\: \: \: \: \: \: \boxed{\tt{ \: \: \: \: \: \: OP \: R_1 \: \to \: R_1 + R_2 + R_3 \: \: \: \: \: }}$

$\rm \: \: \begin{gathered}\sf \left | \begin{array}{ccc}l + m + n&l + m + n&l + m + n\\m&n&l\\n&l& m\end{array}\right | \end{gathered} \: = \: 0$

Take out l + m + n common from first row.

$\rm \: \:(l + m + n) \begin{gathered}\sf \left | \begin{array}{ccc}1&1&1\\m&n&l\\n&l& m\end{array}\right | \end{gathered} \: = \: 0$

$\rm\implies \: \: \:l + m + n = 0 \: \: or \: \begin{gathered}\sf \left | \begin{array}{ccc}1&1&1\\m&n&l\\n&l& m\end{array}\right | \end{gathered} \: = \: 0$

So, Option [B] is correct.

Consider,

$\rm \: \: \begin{gathered}\sf \left | \begin{array}{ccc}1&1&1\\m&n&l\\n&l& m\end{array}\right | \end{gathered} \: = \: 0$

$\: \: \: \: \: \: \boxed{\tt{ \: \: \: \: \: \: OP \: C_2 \: \to \: C_2 - C_1 \: \: \: \: \: }}$

$\rm \: \: \begin{gathered}\sf \left | \begin{array}{ccc}1&0&1\\m&n - m&l\\n&l - n& m\end{array}\right | \end{gathered} \: = \: 0$

$\: \: \: \: \: \: \boxed{\tt{ \: \: \: \: \: \: OP \: C_3 \: \to \: C_ 3- C_1 \: \: \: \: \: }}$

$\rm \: \: \begin{gathered}\sf \left | \begin{array}{ccc}1&0&0\\m&n - m&l - m\\n&l - n& m - n\end{array}\right | \end{gathered} \: = \: 0$

On expanding along Row 1, we get

$\rm \: (n - m)(m - n) - (l - m)(l - n) = 0$

$\rm \: - {(n - m)}^{2} - ( {l}^{2} - ln - lm + mn) = 0$

$\rm \: {(n - m)}^{2} + ( {l}^{2} - ln - lm + mn) = 0$

$\rm \: {n}^{2} + {m}^{2} - 2mn + {l}^{2} - ln - lm + mn = 0$

$\rm \: {n}^{2} + {m}^{2} + {l}^{2} - ln - lm - mn = 0$

$\rm \: 2({n}^{2} + {m}^{2} + {l}^{2} - ln - lm - mn) = 0$

$\rm \: 2{n}^{2} +2 {m}^{2} + 2{l}^{2} -2 ln - 2lm - 2mn = 0$

$\rm \: {n}^{2} + {n}^{2} + {m}^{2} + {m}^{2} + {l}^{2} + {l}^{2} -2 ln - 2lm - 2mn = 0$

$\rm \: {(l - n)}^{2} + {(m - n)}^{2} + {(m - l)}^{2} = 0$

We know, sum of squares is 0 only, when term itself is 0

$\rm\implies \:l - n = 0 \: \: and \: \: n - m = 0 \: \: and \: \: m - n = 0$

$\rm\implies \:l = n \: \: and \: \: n = m\: \: and \: \: m = n$

$\rm\implies \:l = n = m$

So, we concluded that

$\rm\implies \:\boxed{\tt{ \: \: l + m + n = 0 \: \: }}$

Hence, Option [ B ] is correct.

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Additional Information

Different forms of equations of a straight line

1. Equations of horizontal and vertical lines

Equation of line parallel to x - axis passes through the point (a, b) is y = b.

Equation of line parallel to y - axis passes through the point (a, b) is x = a.

2. Point-slope form equation of line

Equation of line passing through the point (a, b) having slope m is y - b = m(x - a)

3. Slope-intercept form equation of line

Equation of line which makes an intercept of c units on y axis and having slope m is y = mx + c.

4. Intercept Form of Line

Equation of line which makes an intercept of a and b units on x - axis and y - axis respectively is x/a + y/b = 1.

5. Normal form of Line

Equation of line which is at a distance of p units from the origin and perpendicular makes an angle β with the positive X-axis is x cosβ + y sinβ = p.