If the lines ax + by + c = 0, bx + cy + a = 0 andcx + ay + b = 0 be concurrent, then

1. Break Down the Problem

We need to determine the condition under which the three lines given by the equations $ax + by + c = 0$, $bx + cy + a = 0$, and $cx + ay + b = 0$ are concurrent. The lines are concurrent if they intersect at a single point.

2. Relevant Concepts

The condition for three lines to be concurrent can be derived from the determinant of their coefficients. For the lines:

• Line 1: $ax + by + c = 0$ can be represented by the coefficients $(a, b, c)$
• Line 2: $bx + cy + a = 0$ can be represented by the coefficients $(b, c, a)$
• Line 3: $cx + ay + b = 0$ can be represented by the coefficients $(c, a, b)$

These lines are concurrent if the following determinant equals zero:

$\begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} = 0$

3. Analysis and Detail

We will calculate the determinant:

$D = \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix}$

Calculating the determinant using the expansion:

$D = a \begin{vmatrix} c & a \\ a & b \end{vmatrix} - b \begin{vmatrix} b & a \\ c & b \end{vmatrix} + c \begin{vmatrix} b & c \\ c & a \end{vmatrix}$

Calculating each of the 2x2 determinants:

1. For $D_1 = \begin{vmatrix} c & a \\ a & b \end{vmatrix} = cb - a^2$

2. For $D_2 = \begin{vmatrix} b & a \\ c & b \end{vmatrix} = bb - ac = b^2 - ac$

3. For $D_3 = \begin{vmatrix} b & c \\ c & a \end{vmatrix} = ba - c^2$

Substituting these back into the determinant $D$:

$D = a(cb - a^2) - b(b^2 - ac) + c(ba - c^2)$

Expanding this leads us to rearranging terms:

$D = acb - a^3 - b^3 + abc + abc - c^3$

Thus:

$D = 3abc - a^3 - b^3 - c^3$

The lines are concurrent if $D = 0$:

4. Verify and Summarize

To summarize, the condition for the lines to be concurrent is:

$3abc - (a^3 + b^3 + c^3) = 0$

Final Answer

The lines $ax + by + c = 0$, $bx + cy + a = 0$, and $cx + ay + b = 0$ are concurrent if:

$3abc = a^3 + b^3 + c^3$