Straight Lines
Straight Line
Allen Star Batch
Grade 11
Question:
Let $P$ be any point on the line $x - y + 3 = 0$ and $A$ be a fixed point $(3,4)$. If the family of lines given by the equation $(3\sec\theta + 5\cos e\theta)x + (7\sec\theta - 3\cos e\theta)y + 11(\sec\theta - \cos e\theta) = 0$ are concurrent at a point $B$ for all permissible values of $\theta$ and maximum value of $|PA - PB| = 2\sqrt{2n}$ ($n \in \mathbb{N}$), then find the value of $n$.
Step-by-Step Solution
Key Concept: A family of lines passing through the intersection of two lines can be represented as a linear combination, and the triangle inequality determines extreme values.
The equation $(3x + 7y + 11)\sec heta + (5x - 3y - 11)\cos e heta = 0$ passes through the intersection of lines $3x + 7y + 11 = 0$ and $5x - 3y - 11 = 0$, which gives point $B = (1, -2)$. Using the triangle inequality, $|PA - PB| \leq AB$ where the maximum value is $|PA - PB|_{\max} = AB = \sqrt{6^2 + 2^2} = \sqrt{40} = 2\sqrt{10}$, hence $n = 5$.
Correct Answer: 5