Animated Solution for Mathematics - Straight Lines: Let $A(1,2)$ and $C(-3,-6)$ be two diagonally opposite vertices of a rhombus, whose sides AD and BC are parallel to the line $7x-y=14$. If $B(\alpha ,\beta )$ and $D(\gamma ,\delta )$ are the other two vertices, then $|\alpha +\beta +\gamma +\delta |$ is equal to

Welcome, fellow traveler of the coordinate plane. Today, we are not just solving a problem; we are exploring the elegant architecture of a rhombus.

A rhombus is a creature of perfect symmetry, and in the world of JEE Advanced, symmetry is your greatest ally. When you see a problem involving a rhombus, do not just see coordinates—see the balance, the perpendicularity, and the intersection of constraints.

We begin with two points, $A(1,2)$ and $C(-3,-6)$. These are the endpoints of one diagonal.

In a rhombus, the diagonals are the 'skeleton' of the shape. They bisect each other, meaning they meet at their mutual midpoint, $M$. This point $M$ is the center of gravity for our entire figure.

By finding $M$, we have anchored our entire construction. Everything else will revolve around this point.

Now, we must find the other diagonal, $BD$. We know the slope of $AC$ is $m_1=\frac{-6-2}{-3-1}=2$.

The diagonals of a rhombus are perpendicular. This is a gift! It means the slope of $BD$, let's call it $m_2$, must satisfy the condition $m_1\cdot m_2=-1$. Thus, $m_2=-\frac{1}{2}$.

With a point $M(-1,-2)$ and a slope of $-\frac{1}{2}$, we can write the equation of the line $BD$ using the point-slope form:

This line is the path upon which vertices $B$ and $D$ must lie. We have constrained the infinite plane down to a single line.

We are given that side $AD$ is parallel to the line $7x-y=14$. Parallel lines share the same slope. By rewriting the given line as $y=7x-14$, we see the slope is $7$.

Since $AD$ passes through $A(1,2)$, its equation is:

Vertex $D$ is the intersection of side $AD$ and diagonal $BD$. We substitute the equation of $AD$ into the equation of $BD$:

Substituting $x=\frac{1}{3}$ back into $y=7x-5$, we find $y=-\frac{8}{3}$. Thus, $D=(\frac{1}{3},-\frac{8}{3})$.

We have found $D$, but we need $B$. We use the midpoint property. Since $M(-1,-2)$ is the midpoint of $BD$, we know:

Using our values for $\gamma =\frac{1}{3}$ and $\delta =-\frac{8}{3}$, we solve for $\alpha$ and $\beta$:

The final step is to calculate the absolute value of the sum of the coordinates of $B$ and $D$:

The final result is 6. Through the simple application of geometric properties and algebraic precision, we have unraveled the rhombus.