Question

A rectangle is formed by the lines \( x = 0 \), \( y = 0 \), \( x = 3 \) and \( y = 4 \). Let the line \( L \) be perpendicular to \( 3x + y + 6 = 0 \) and divide the area of the rectangle into two equal parts. Then the distance of the point \( \left(\frac{1}{2}, -5\right) \) from the line \( L \) is equal to :

• A. \( 2\sqrt{5} \)
• B. \( 2\sqrt{10} \)
• C. \( 3\sqrt{10} \)
• D. \( \sqrt{10} \)

Hint:

Any line passing through the center of a rectangle, parallelogram, or circle bisects its area.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
A line that divides the area of a rectangle into two equal parts must pass through the center of the rectangle.
Step 2: Key Formula or Approach:
1. Center of rectangle: \( (\frac{0+3}{2}, \frac{0+4}{2}) = (1.5, 2) \).
2. Slope of given line \( 3x+y+6=0 \) is \( -3 \).
3. Slope of perpendicular line \( L \) is \( m = 1/3 \).
Step 3: Detailed Explanation:
Equation of line \( L \): \[ y - 2 = \frac{1}{3}(x - 1.5) \implies 3y - 6 = x - 1.5 \implies x - 3y + 4.5 = 0 \] Multiply by 2 for integer constants: \( 2x - 6y + 9 = 0 \). Distance from \( (1/2, -5) \): \[ d = \left| \frac{2(1/2) - 6(-5) + 9}{\sqrt{2^2 + (-6)^2}} \right| = \left| \frac{1 + 30 + 9}{\sqrt{40}} \right| = \frac{40}{\sqrt{40}} = \sqrt{40} = 2\sqrt{10} \]
Step 4: Final Answer:
The distance is \( 2\sqrt{10} \).