(B) If the point $P$ is symmetric to the point $Q(4, -1)$ with respect to the bisector of the first quadrant, then the length of $PQ$ is:
Step-by-Step Solution
Key Concept: Use AM-GM inequality on the sum of intercepts to find the minimum value, and apply reflection and distance formulas for geometric configurations.
For part (A), the line equation is $y - 4 = m(x - 1)$ with $m < 0$. We find $OA = 1 + rac{4}{-m}$ and $OB = 4 + (-m)$. Applying AM ≥ GM on $ rac{4}{-m}$ and $(-m)$, we get $ rac{ rac{4}{-m} + (-m)}{2} \geq \sqrt{ rac{4}{-m} \cdot (-m)} = 2$, which gives $OA + OB \geq 9$. For part (B), the image of $Q(4, -1)$ with respect to line mirror $y = x$ is $P(-1, 4)$, so $PQ = \sqrt{50} = 4\sqrt{2}$. For part (C), from the figure showing a rhombus with vertices at $(0, 2)$, $(2, 0)$, and symmetric points, the perpendicular distances from origin $O$ to sides $AB$, $CD$, $AD$, and $BC$ are all $\sqrt{2}$. For part (D), the line equation $x = 4 + rac{\lambda}{\sqrt{2}}$ and $y = -1 + \sqrt{2}\lambda$ simplifies to $2x - y = 9$, giving x-intercept $ rac{9}{2}$.
Correct Answer: [A-s] [B-p] [C-q] [D-r]