Question

Let the equations of two adjacent sides of a parallelogram ABCD be 2x – 3y = –23 and 5x + 4y = 23. If the equation of its one diagonal AC is 3x + 7y = 23 and the distance of A from the other diagonal is d, then 50 d2 is equal to____.

Answer 1

Correct Answer: 529

To solve the problem, we need to determine the distance of point A from the other diagonal of the parallelogram and evaluate 50 d². Let's proceed step by step:

1. Identify the coordinates of A: We find the intersection of the lines 2x – 3y = –23 and 3x + 7y = 23.

   Solving simultaneously:
   Multiply 2x – 3y = –23 by 3:
   6x – 9y = –69 (Equation 1)
   Multiply 3x + 7y = 23 by 2:
   6x + 14y = 46 (Equation 2)

   Subtract Equation 1 from Equation 2:
   (6x + 14y) – (6x – 9y) = 46 + 69
   23y = 115
   y = 5

   Substitute y = 5 into 2x – 3y = –23:
   2x – 3(5) = –23
   2x – 15 = –23
   2x = –8
   x = –4
   Thus, A = (–4, 5).

2. Equation of the other diagonal BD: The side parallel to 5x + 4y = 23 through A is 5x + 4y = k.
   Substitute A's coordinates: 5(–4) + 4(5) = k
   k = –20 + 20 = 0
   Thus, 5x + 4y = 0 is the equation of BD.

3. Calculate the distance from A to BD: Using the point-to-line distance formula:
   d = |Ax₁ + By₁ + C| / √(A² + B²)
   For 5x + 4y = 0, A = 5, B = 4, C = 0, (x₁, y₁) = (–4, 5)
   d = |5(–4) + 4(5) + 0| / √(5² + 4²)
   d = |–20 + 20| / √(25 + 16)
   d = 0 / √41 = 0

4. Find 50 d²: Since d = 0, 50d² = 50(0)² = 0.

Thus, 50 d² = 0, which falls outside the given range of [529, 529]. There might be an issue in the computation or problem statement.