A variable line '$L$' of the form $y = mx$ is drawn to meet the lines $L_1: 2x + 3y - 5 = 0$; $L_2: x + 2y - 5 = 0$ and $L_3: 6x + 4y - 5 = 0$ at points $A$, $B$ and $C$. A point $P(a,b)$ is taken on the line '$L$' also $\frac{k(a+b)}{OP} = \frac{1}{OA} + \frac{1}{OB} + \frac{1}{OC}$ then value of $k$ is ____.
Step-by-Step Solution
Key Concept: The sum of reciprocals of radii from concurrent lines through origin relates to the harmonic property of collinear points.
Given three points $A$, $B$, $C$ on lines $L_1$, $L_2$, $L_3$ respectively, where each line passes through the origin with slopes determined by the constraint $2r_1\cos\theta + 3r_1\sin\theta - 5 = 0$. Using the relationship $\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{9}{5}(\cos\theta + \sin\theta)$, we find $k = \frac{9}{5}$ by expressing the sum in terms of the dot product with direction vector $(a+b)$.
Correct Answer: 1.80