Question

If the straight lines $ax+may+1=0, bx+(m+1)by+1=0$ and $cx+(m+2)cy+1=0$, (where a, b, c and m are non-zero) are concurrent, then a,b,c are in-

• A. A.P. only for m = 1
• B. AP for all m
• C. GP for all m
• D. HP for all m

Answer 1

Answer: (D)

HP for all m

Answer 2

The condition for three lines $A_1x + B_1y + C_1 = 0$, $A_2x + B_2y + C_2 = 0$, and $A_3x + B_3y + C_3 = 0$ to be concurrent is that the determinant of their coefficients is zero:

$\begin{vmatrix} A_1 & B_1 & C_1 \\ A_2 & B_2 & C_2 \\ A_3 & B_3 & C_3 \end{vmatrix} = 0$

Substituting the coefficients from the given lines:

$\begin{vmatrix} a & ma & 1 \\ b & (m+1)b & 1 \\ c & (m+2)c & 1 \end{vmatrix} = 0$

Applying column operation $C_2 \to C_2 - m C_1$:

$\begin{vmatrix} a & 0 & 1 \\ b & b & 1 \\ c & 2c & 1 \end{vmatrix} = 0$

Expanding the determinant:

$a(b - 2c) + 1(2bc - bc) = 0$ $ab - 2ac + bc = 0$ $ab + bc = 2ac$

Dividing by $abc$:

$\frac{1}{c} + \frac{1}{a} = \frac{2}{b}$

This shows that $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in Arithmetic Progression (AP), thus a, b, c are in Harmonic Progression (HP). Since the relationship is independent of 'm', a, b, c are in HP for all values of m.