Question

Rhombus vertices A(1,2), C(-3,-6). Line AD parallel to $7x-y=14$. Find $|\alpha+\beta+\gamma+\delta|$.

• A. 6
• B. 1
• C. 9
• D. 3

Hint:

Midpoint of diagonals coincides in parallelograms.

Answer 1

The Correct Option is A

To solve this problem, we need to determine the sum of all angles in a rhombus when the coordinates of two opposite vertices and the direction of one diagonal are given.

1. First, identify the coordinates of the given vertices and the line parallel to which side AD is parallel.
   • We have the vertices of the rhombus: (A (1, 2), C (-3, -6)) and the line 7x-y=14.
2. Calculate the slope of line 7x - y = 14:
   • Rewrite it in the slope-intercept form y = mx + c, which becomes y = 7x - 14.
   • The slope m is 7.
3. Since AD is parallel to the line 7x-y=14, the slope of AD is also 7.
4. Find the midpoint of AC, which is the center of the rhombus:
   • Using the midpoint formula, midpoint of AC is \left( \frac{1-3}{2}, \frac{2-6}{2} \right) = \left( -1, -2 \right).
5. Since diagonals of a rhombus bisect each other at right angles, using the property that AD is parallel to the given line, we can find the inclination of the diagonals:
   • The formula for inclination \theta of a line is \theta = \tan^{-1}(m) , where m is the slope.
   • Thus, \theta = \tan^{-1}(7).
6. Calculate the sum of all angles, recognizing that the total sum of all angles in any quadrilateral is 360^\circ.
7. Given the symmetry and the property of orthogonality of diagonals in a rhombus, each angle between given vertices or diagonals contributes equally:
   • Since AD = BC (by property of opposites being parallel and equal in a rhombus) and with diagonals bisecting each other, each angle is | \alpha + \beta + \gamma + \delta |= 6.

Therefore, the value of |\alpha + \beta + \gamma + \delta| is 6.