Question

The centre of a square of side 4 units length is \( (3, 7) \), and one of the diagonals is parallel to the line \( y = x \). If \( (x_1, y_1), (x_2, y_2), (x_3, y_3), (x_4, y_4) \) are the vertices of this square, then \[ \frac{y_1 y_2 y_3 y_4}{x_1 x_2 x_3 x_4} = ? \]

• A. \( 81 \)
• B. \( \frac{245}{16} \)
• C. \( 25 \)
• D. \( \frac{105}{2} \)

Hint:

Use symmetry and rotation properties of squares to compute vertices from the center and diagonal direction.

Answer 1

The Correct Option is A

Given center: \( C = (3, 7) \), side = 4 ⇒ half-diagonal = \( \frac{4}{\sqrt{2}} = 2\sqrt{2} \) Since diagonal is along \( y = x \), the direction vector of the diagonal is \( \langle 1, 1 \rangle \), unit vector = \( \frac{1}{\sqrt{2}}(1,1) \) Endpoints of one diagonal: \[ A = (3 + 2, 7 + 2), B = (3 - 2, 7 - 2) = (5, 9), (1, 5) \] Other diagonal is perpendicular to this (along \( y = -x \)), direction vector \( \langle 1, -1 \rangle \) Use same magnitude to get the other two points: \[ C = (3 + 2, 7 - 2) = (5, 5), D = (1, 9) \] Vertices: \( (5, 9), (1, 5), (5, 5), (1, 9) \) Now compute: \[ \frac{y_1 y_2 y_3 y_4}{x_1 x_2 x_3 x_4} = \frac{9 \cdot 5 \cdot 5 \cdot 9}{5 \cdot 1 \cdot 5 \cdot 1} = \frac{2025}{25} = 81 \]