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

The equations of the adjacent sides are: \[ 2x - 3y = -23 \quad \text{(1)} \] \[ 5x + 4y = 23 \quad \text{(2)} \] Now, solve for the coordinates of the points \( A \) and \( C \). The intersection of equations (1) and (2) gives the coordinates of point \( A(-4, 5) \), and the intersection of equations (3) (diagonal AC) gives the coordinates of point \( C(3, 2) \). Now, we find the midpoint of diagonal \( AC \): \[ \text{Midpoint of AC} = \left( \frac{-4 + 3}{2}, \frac{5 + 2}{2} \right) = \left( -\frac{1}{2}, \frac{7}{2} \right) \] Next, we need to find the equation of the diagonal \( BD \). The midpoint of \( BD \) is the same as the midpoint of \( AC \), and using the equation of the line passing through points \( B(-1, -7) \) and \( D(1, 2) \), we obtain: \[ \frac{y - \frac{7}{2}}{x + \frac{1}{2}} = \frac{\frac{7}{2} - 2}{-\frac{1}{2} - 1} \] \[ y - \frac{7}{2} = \frac{\frac{7}{2} - 2}{-\frac{1}{2} - 1} \left( x + \frac{1}{2} \right) \] \[ 7x + y = 0 \] Now, the distance of point \( A(-4, 5) \) from the diagonal \( BD \) is calculated using the formula for the distance from a point to a line: \[ d = \frac{|7(-4) + 5|}{\sqrt{7^2 + 1^2}} = \frac{| -28 + 5 |}{\sqrt{49 + 1}} = \frac{| -23 |}{\sqrt{50}} = \frac{23}{\sqrt{50}}. \] Now calculate \( 50d^2 \): \[ 50d^2 = 50 \times \left( \frac{23}{\sqrt{50}} \right)^2 = 50 \times \frac{529}{50} = 529. \] Thus, \( 50d^2 = 529 \).