Let A(-2, -1), B(1, 0), C(α, β) and D(γ, δ) be the vertices of a parallelogram ABCD. If the point C

Q: Why can't we just use the slope property of parallel sides?

You absolutely can! The slope of $AB$ must equal the slope of $DC$. However, using the midpoint property is often more elegant because it avoids dealing with fractions and potential division-by-zero errors. In JEE Advanced, choosing the path of least resistance is a skill in itself.

Q: Wait, why do we take the absolute value at the end?

The question specifically asks for $|\alpha + \beta + \gamma + \delta|$. In coordinate geometry, coordinates can be negative, leading to a negative sum. The absolute value ensures we are looking at the magnitude of the result, which is a common way to ensure a positive integer answer in numerical-type questions.

Q: What if the lines were parallel?

If the lines $2x - y = 5$ and $3x - 2y = 6$ were parallel, they would never intersect, and the parallelogram would not be uniquely defined. Always check the slopes of the constraint lines first to ensure the problem is well-posed!

Question

Let $A(-2, -1), B(1, 0), C(\alpha, \beta)$ and $D(\gamma, \delta)$ be the vertices of a parallelogram $ABCD$. If the point $C$ lies on $2x - y = 5$ and the point $D$ lies on $3x - 2y = 6$, then the value of $|\alpha + \beta + \gamma + \delta|$ is equal to _______.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a coordinate plane, looking at a parallelogram $ABCD$. You know where $A$ and $B$ are, but $C$ and $D$ are elusive, hiding on specific lines.

In any parallelogram, the diagonals $AC$ and $BD$ are not just segments; they are the axes of symmetry that bisect each other. This is the secret key to the entire problem.

By recognizing that the midpoint of $AC$ must be identical to the midpoint of $BD$, we instantly bridge the gap between the known and the unknown.

### Translating Geometry into Algebra

Let us formalize this. The midpoint of $AC$ is given by the average of its coordinates:

$$M_{AC} = \left( \frac{\alpha - 2}{2}, \frac{\beta - 1}{2} \right)$$

Similarly, the midpoint of $BD$ is:

$$M_{BD} = \left( \frac{\gamma + 1}{2}, \frac{\delta + 0}{2} \right)$$

Because these two points are the same, we can equate their components. This gives us two beautiful, simple relationships: $\gamma = \alpha - 3$ and $\delta = \beta - 1$.

We have effectively reduced the number of variables from four to two. We are no longer chasing four ghosts; we are chasing two.

### The Constraint Trap

Now, we must respect the constraints. Point $C(\alpha, \beta)$ is bound to the line $2x - y = 5$, which translates to $2\alpha - \beta = 5$.

Point $D(\gamma, \delta)$ is bound to $3x - 2y = 6$. This is where many students stumble, but you will not.

We substitute our expressions for $\gamma$ and $\delta$ into the second line equation:

$$3(\alpha - 3) - 2(\beta - 1) = 6$$

Expanding this, we get $3\alpha - 9 - 2\beta + 2 = 6$, which simplifies to the following linear constraint:

$$3\alpha - 2\beta = 13$$

### The Final Convergence

We now have a system of two linear equations:

$$2\alpha - \beta = 5$$
$$3\alpha - 2\beta = 13$$

From the first, we know $\beta = 2\alpha - 5$. Substituting this into the second, we get:

$$3\alpha - 2(2\alpha - 5) = 13$$

This simplifies to $3\alpha - 4\alpha + 10 = 13$, leading us to $\alpha = -3$.

With $\alpha$ in hand, the rest falls like dominoes: $\beta = -11$, $\gamma = -6$, and $\delta = -12$.

The sum is $-3 - 11 - 6 - 12 = -32$. Taking the absolute value, we arrive at **32**.

You have successfully navigated the constraints and unlocked the geometry. This is the power of coordinate geometry: turning a visual puzzle into a logical, solvable path.

Enter Numerical Value:

Visualized Solution

Visualizing the Parallelogram $A B C D$

The Midpoint Property

Equating the $x$ -coordinates

Equating the $y$ -coordinates

Using the Line Constraint for $C$

Using the Line Constraint for $D$

Simplifying the Equation for $D$

Solving for $α$

Finding $β, γ,$ and $δ$

Calculating the Final Sum

The Sigma Insight: Distance and Section Formulas

Analyzing the Setup

Translating Geometry into Algebra

The Constraint Trap

The Final Convergence

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Enter Numerical Value:

Visualized Solution

Visualizing the Parallelogram .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ABCD

The Midpoint Property

Equating the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x-coordinates

Equating the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y-coordinates

Using the Line Constraint for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Using the Line Constraint for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D

Simplifying the Equation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } D

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β,γ, and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } δ