Question

The combined equation of a possible pair of adjacent sides of a square with area 16 square units whose centre is the point of intersection of the lines \[ x + 2y - 3 = 0 \quad \text{and} \quad 2x - y - 1 = 0 \] is: \

• A. \( (2x - y - 1 + 4\sqrt{5})(x + 2y - 3 + 4\sqrt{5}) = 0 \)
• B. \( (2x - y - 1 - 4\sqrt{5})(x + 2y - 3 - 4\sqrt{5}) = 0 \)
• C. \( (2x - y - 2\sqrt{5})(x + 2y + 2\sqrt{5}) = 0 \)
• D. \( (2x - y - 1 - 2\sqrt{5})(x + 2y - 3 + 2\sqrt{5}) = 0 \)
  \

Hint:

For squares with known diagonals, find the center and perpendicular sides using their slopes. Translate by the required distance to get the final equation.

Answer 1

The Correct Option is D

Step 1: Finding the Center of the Square
The given lines: \[ x + 2y - 3 = 0, \quad 2x - y - 1 = 0 \] represent diagonals of the square. The center of the square is their intersection. Solving for \( x \) and \( y \): Multiplying the first equation by 2: \[ 2x + 4y - 6 = 0. \] Subtracting the second equation: \[ (2x + 4y - 6) - (2x - y - 1) = 0. \] \[ 5y - 5 = 0. \] \[ y = 1. \] Substituting \( y = 1 \) into \( x + 2y - 3 = 0 \): \[ x + 2(1) - 3 = 0. \] \[ x = 1. \] Thus, the center of the square is \( (1,1) \). Step 2: Determining the Length of the Side
The area of the square is given as 16, so the side length is: \[ s = \sqrt{16} = 4. \] Step 3: Finding the Required Pair of Perpendicular Sides
Lines perpendicular to the diagonals have slopes: - Given diagonal slopes: \( -\frac{1}{2} \) and \( 2 \). - Perpendicular slopes: \( 2 \) and \( -\frac{1}{2} \). Using point-slope form \( y - y_1 = m(x - x_1) \) at \( (1,1) \): 1. For slope \( 2 \): \[ y - 1 = 2(x - 1). \] \[ y = 2x - 2 + 1. \] \[ 2x - y - 1 = 0. \] 2. For slope \( -\frac{1}{2} \): \[ y - 1 = -\frac{1}{2} (x - 1). \] \[ y = -\frac{1}{2}x + \frac{1}{2} + 1. \] \[ y = -\frac{1}{2}x + \frac{3}{2}. \] Multiplying by 2: \[ x + 2y - 3 = 0. \] Step 4: Finding Shifted Equations
To get equations of the square sides, shift by \( \pm 2\sqrt{5} \): \[ (2x - y - 1 - 2\sqrt{5})(x + 2y - 3 + 2\sqrt{5}) = 0. \] Step 5: Conclusion
Thus, the final answer is: \[ \boxed{(2x - y - 1 - 2\sqrt{5})(x + 2y - 3 + 2\sqrt{5}) = 0}. \] \bigskip