The point (4, 1) undergoes the following three transformations successively.
(i) Reflection about th

Q: Does the order of transformations matter?

Absolutely! Transformations are like a recipe. If you bake before you mix, you get a disaster. In linear algebra, matrix multiplication is non-commutative, meaning $AB \neq BA$. Always follow the sequence given in the problem to ensure you arrive at the correct final position.

Q: Why is the rotation formula $x \cos \theta - y \sin \theta$ and not something else?

This formula is derived from the geometry of a point $(r, \phi)$ rotating to $(r, \phi + \theta)$. By using the trigonometric addition formulas for sine and cosine, we can express the new coordinates in terms of the original $x$ and $y$. It is the fundamental bedrock of rotational mechanics in coordinate geometry.

Q: Can I use matrices for all of these?

Yes, and it is a very elegant approach! You can represent each transformation as a matrix: a reflection matrix $M_1$, a translation matrix $M_2$, and a rotation matrix $M_3$. The final point is $M_3(M_2(M_1 P_0))$. It is a powerful, systematic way to handle multiple transformations without getting lost in the algebra.

Question

The point $(4, 1)$ undergoes the following three transformations successively. 
(i) Reflection about the line $y = x$. 
(ii) Translation through a distance 2 units along the positive direction of x-axis. 
(iii) Rotation through an angle $\pi/4$ about the origin in the counter clockwise direction. 
Then the final position of the point is given by the coordinates.

Options:

(A) $(\frac{1}{\sqrt{2}}, \frac{7}{\sqrt{2}})$

(B) $(-\sqrt{2}, 7\sqrt{2})$

(C) $(-\frac{1}{\sqrt{2}}, \frac{7}{\sqrt{2}})$

(D) $(\sqrt{2}, 7\sqrt{2})$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the mathematical realm. Today, we are not just solving a problem; we are choreographing a dance.

Imagine a point, $P_0(4, 1)$, standing on a vast, blank coordinate plane. It is about to undergo a series of transformations—a reflection, a translation, and a rotation—that will shift its existence across the grid.

This is a classic JEE Advanced challenge, and it tests not just your ability to calculate, but your ability to visualize the geometry behind the algebra.

### The Mirror of Symmetry

Our journey begins with a reflection about the line $y = x$. Think of this line as a perfect, diagonal mirror slicing through the origin.

When a point is reflected across this mirror, it does not just move; it swaps its identity. The rule is elegant and simple: $(x, y) 	o (y, x)$.

Our point $P_0(4, 1)$ steps into this mirror and emerges as $P_1(1, 4)$. It is a clean, swift change, a reminder that in coordinate geometry, symmetry is often the shortest path to clarity.

### The Horizontal Slide

With our point now at $P_1(1, 4)$, we move to the second act: translation. Translation is the simplest of movements—a pure, unadulterated slide.

We are told to move the point $2$ units along the positive direction of the $x$-axis. This means our $y$-coordinate remains untouched, a silent observer, while our $x$-coordinate grows.

We take $x = 1$ and add $2$, landing us at $x = 3$. Our point, now $P_2(3, 4)$, has successfully navigated the slide. It is a moment of stability before the final, more complex maneuver.

### The Grand Finale: Rotation

Now, we reach the climax: a rotation through an angle $	heta = \frac{\pi}{4}$ about the origin in the counter-clockwise direction. This is where many students stumble, but let us approach it with calm precision.

We use the rotation formulas: 
$$x' = x \cos 	heta - y \sin 	heta$$
$$y' = x \sin 	heta + y \cos 	heta$$

These are not just arbitrary equations; they are the mathematical embodiment of a spin. With $	heta = \frac{\pi}{4}$, we know that $\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$ and $\sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$.

Substituting our current coordinates $(3, 4)$ into these formulas, we calculate:
$$x' = 3\left(\frac{1}{\sqrt{2}}ight) - 4\left(\frac{1}{\sqrt{2}}ight) = \frac{3 - 4}{\sqrt{2}} = -\frac{1}{\sqrt{2}}$$

Similarly, for the $y$-coordinate:
$$y' = 3\left(\frac{1}{\sqrt{2}}ight) + 4\left(\frac{1}{\sqrt{2}}ight) = \frac{3 + 4}{\sqrt{2}} = \frac{7}{\sqrt{2}}$$

### The Final Destination

We have arrived. After the reflection, the slide, and the spin, our point $P_3$ rests at:
$$\left(-\frac{1}{\sqrt{2}}, \frac{7}{\sqrt{2}}ight)$$

It is a beautiful result, a testament to the power of breaking down a complex problem into manageable, logical steps.

Remember, in JEE Advanced, the complexity is often just a mask for a sequence of simple, fundamental truths. Keep practicing, keep visualizing, and most importantly, keep falling in love with the elegance of the math.

Select Answer:

Visualized Solution

$I ni t ia l P o in t P_{0} (4, 1)$

$T r an s f or ma t i o n 1 : R e f l ec t i o n$

$A ppl y in g t h e R e f l ec t i o n$

$T r an s f or ma t i o n 2 : T r an s l a t i o n$

$A ppl y in g t h e T r an s l a t i o n$

$T r an s f or ma t i o n 3 : R o t a t i o n S e t u p$

$S u b s t i t u t in g V a l u es f or R o t a t i o n$

$C a l c u l a t in g t h e N e w x - coor d ina t e$

$C a l c u l a t in g t h e N e w y - coor d ina t e$

$F ina l P os i t i o n$

The Sigma Insight: Distance and Section Formulas

Analyzing the Setup

The Mirror of Symmetry

The Horizontal Slide

The Grand Finale

The Final Destination

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } InitialPointP0​(4,1)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Transformation1:Reflection

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Transformation2:Translation

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Transformation3:RotationSetup

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } CalculatingtheNewx−coordinate

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } CalculatingtheNewy−coordinate

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