A square of side a lies above the x-axis and has one vertex at the origin. The side passing through

Q: Wait, why is the x-coordinate of $C$ negative? It's a square, shouldn't all coordinates be positive?

That is a classic trap! Remember, the square is tilted. Because the side $OC$ is perpendicular to $OA$, it is rotated by $90^\circ + \alpha$. Since $0 < \alpha < \frac{\pi}{4}$, the angle of $OC$ is between $90^\circ$ and $135^\circ$, placing it in the second quadrant. Hence, the x-coordinate must be negative. Trust the trigonometry!

Q: Is there a faster way to find the slope without doing all this algebra?

You could use vector rotation matrices, but for a JEE problem, the coordinate geometry approach is the most reliable. The algebra might look heavy, but it is designed to cancel out beautifully. The 'fast' way is to be confident in your algebraic manipulation!

Q: I got stuck during the expansion of the line equation. How do I know if I'm on the right track?

If you see terms like $a \sin \alpha \cos \alpha$ appearing on both sides of your equation, you are doing it perfectly! Those terms are meant to cancel out. If they don't, re-check your signs during the cross-multiplication step.

Question

A square of side $a$ lies above the x-axis and has one vertex at the origin. The side passing through the origin makes an angle $\alpha$ ($0 < \alpha < \frac{\pi}{4}$) with the positive direction of x-axis. The equation of its diagonal not passing through the origin is

Options:

(A) $y(\cos \alpha + \sin \alpha) + x(\cos \alpha - \sin \alpha) = a$

(B) $y(\cos \alpha - \sin \alpha) - x(\sin \alpha - \cos \alpha) = a$

(C) $y(\cos \alpha + \sin \alpha) + x(\sin \alpha - \cos \alpha) = a$

(D) $y(\cos \alpha + \sin \alpha) + x(\sin \alpha + \cos \alpha) = a$

Sigma Percentile · Accepted Answer

### The Geometry of the Tilted Square

Welcome, future engineer. Today, we are not just solving a coordinate geometry problem; we are embarking on a journey of visualization.

Imagine you are standing at the origin of a Cartesian plane. In front of you lies a square, $OABC$, with side length $a$. But this isn't a boring, axis-aligned square; it is tilted, resting on a slope of angle $\alpha$.

If you can master the geometry of this rotation, you can master any problem in JEE Advanced.

### Phase 1: The Art of Coordinates

Let us begin by pinning down our vertices. We know vertex $O$ is at $(0,0)$.

The side $OA$ makes an angle $\alpha$ with the positive x-axis. This is a classic application of polar coordinates. If the length of the side is $a$, then the coordinates of $A$ are simply $(a \cos \alpha, a \sin \alpha)$.

Now, consider vertex $C$. In a square, the sides meeting at the origin are perpendicular. This means the side $OC$ is at an angle of $90^{\circ} + \alpha$ from the positive x-axis.

Using our trigonometric definitions, the coordinates of $C$ become $(a \cos(90^{\circ} + \alpha), a \sin(90^{\circ} + \alpha))$. Applying the reduction formulas, we find $C$ is at $(-a \sin \alpha, a \cos \alpha)$.

Do you see the beauty here? The negative sign in the x-coordinate is not an error; it is the physical manifestation of the square rotating into the second quadrant. Always trust the math over your intuition when the geometry gets complex.

### Phase 2: The Slope of the Diagonal

We need the equation of the diagonal $AC$. To find the equation of any line, we need two things: a point and a slope.

We have the points $A(a \cos \alpha, a \sin \alpha)$ and $C(-a \sin \alpha, a \cos \alpha)$. Let us calculate the slope $m$ using the standard formula $m = \frac{y_2 - y_1}{x_2 - x_1}$.

Substituting our coordinates, we get:

$$m = \frac{a \cos \alpha - a \sin \alpha}{-a \sin \alpha - a \cos \alpha}$$

It looks messy, doesn't it? But look closer. We can factor out the $a$ from the numerator and denominator and cancel it.

Then, pull out a negative sign from the denominator. We are left with:

$$m = \frac{\cos \alpha - \sin \alpha}{-(\sin \alpha + \cos \alpha)}$$

This is the slope of our diagonal. It captures the essence of the tilt $\alpha$.

### Phase 3: The Algebraic Dance

Now, we use the point-slope form: $y - y_1 = m(x - x_1)$. Let us use point $A$ as our anchor.

The equation becomes:

$$y - a \sin \alpha = \frac{\cos \alpha - \sin \alpha}{-(\sin \alpha + \cos \alpha)} (x - a \cos \alpha)$$

This is where most students panic. The expression looks daunting, but take a deep breath. We are going to cross-multiply.

Multiply both sides by $-(\sin \alpha + \cos \alpha)$:

$$-(y - a \sin \alpha)(\sin \alpha + \cos \alpha) = (x - a \cos \alpha)(\cos \alpha - \sin \alpha)$$

Now, expand both sides carefully. On the left, we get $-y(\sin \alpha + \cos \alpha) + a \sin^2 \alpha + a \sin \alpha \cos \alpha$. On the right, we get $x(\cos \alpha - \sin \alpha) - a \cos^2 \alpha + a \sin \alpha \cos \alpha$.

Do you see it? The term $a \sin \alpha \cos \alpha$ appears on both sides. It is a gift! It cancels out perfectly. This is the elegance of mathematics—the complexity vanishes when you follow the logic.

### Phase 4: The Grand Finale

After the cancellation, we group the variables on one side and the constants on the other:

$$y(\cos \alpha + \sin \alpha) + x(\cos \alpha - \sin \alpha) = a \sin^2 \alpha + a \cos^2 \alpha$$

Finally, factor out the $a$ on the right side. We are left with $a(\sin^2 \alpha + \cos^2 \alpha)$.

We know the fundamental identity $\sin^2 \alpha + \cos^2 \alpha = 1$. The entire right side collapses into just $a$.

Our final equation is:

$$y(\cos \alpha + \sin \alpha) + x(\cos \alpha - \sin \alpha) = a$$

You have successfully navigated the rotation, the coordinate geometry, the slope calculation, and the algebraic simplification. This is how you conquer JEE Advanced problems—one logical step at a time. Keep this confidence, and you will be unstoppable.

Select Answer:

Visualized Solution

Visualizing the Square $O A B C$

Coordinates of Vertex $A$

Coordinates of Vertex $C$

Identifying the Target Diagonal

Slope of Diagonal $A C$

Simplifying the Slope

Point-Slope Form Setup

Cross-Multiplication

Expansion of Terms

Grouping and Cancellation

The Final Equation

The Sigma Insight: Various Forms of Equations of a Line

The Geometry of the Tilted Square

Phase 1

Phase 2

Phase 3

Phase 4

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