A man starts walking from the point P(-3, 4), touches the x-axis at R, and then turns to reach at th

Q: Wait, why do we reflect $Q$ and not $P$? Does it change the answer?

That is a fantastic question! The beauty of the Reflection Principle is that it is perfectly symmetric. You could absolutely reflect $P(-3, 4)$ across the $x$-axis to get $P'(-3, -4)$ and then find the line connecting $P'$ and $Q$. You would arrive at the exact same point $R$ on the $x$-axis. The geometry is invariant, so choose whichever point feels more intuitive to you!

Q: Why can't we just use calculus to minimize the distance function $f(x) = \sqrt{(x+3)^2 + 16} + \sqrt{x^2 + 4}$?

You absolutely can! If you differentiate that function with respect to $x$ and set it to zero, you will get the correct answer. However, in JEE Advanced, time is your most precious resource. The Reflection Principle allows you to solve this in seconds using simple coordinate geometry, bypassing the need for complex derivatives and algebraic manipulation. Always look for the geometric shortcut!

Q: Is the path always a straight line? What if the man had to touch the $x$-axis at two different points?

In this specific problem, yes, the shortest path is a straight line because we are in a Euclidean plane. If the problem were more complex—say, requiring the man to touch the $x$-axis at two points—the path would become a series of straight-line segments, and we would apply the reflection principle multiple times. The core principle remains: minimize the distance by 'unfolding' the path into a straight line.

Question

A man starts walking from the point $P(-3, 4)$, touches the $x$-axis at $R$, and then turns to reach at the point $Q(0, 2)$. The man is walking at a constant speed. If the man reaches the point $Q$ in the minimum time, then $50((PR)^2 + (RQ)^2)$ is equal to .

Sigma Percentile · Accepted Answer

### The Geometry of the Shortest Path: A Journey of Optimization

Imagine you are standing at point $P(-3, 4)$ on a coordinate plane. You have a mission: you must touch the $x$-axis at some point $R$ and then proceed to your destination at $Q(0, 2)$.

You are walking at a constant speed, which means that to reach your destination in the minimum time, you must minimize the total distance traveled. This is a classic optimization challenge that has fascinated mathematicians for centuries.

### Phase 1: The Reflection Principle

At first glance, you might be tempted to write down the distance formula for $PR$ and $RQ$, sum them up, and dive into a messy calculus derivative. Stop! Take a breath.

In JEE Advanced, we look for elegance. We use the **Reflection Principle**. Imagine the $x$-axis is a mirror. If you look at point $Q(0, 2)$ in this mirror, you see an image, $Q'(0, -2)$, located exactly on the other side.

Because of the symmetry of reflection, for any point $R$ on the $x$-axis, the distance $RQ$ is identical to the distance $RQ'$. We are no longer trying to minimize $PR + RQ$; we are now trying to minimize $PR + RQ'$.

### Phase 2: The Straight Line

Why is this transformation so powerful? Because the shortest distance between any two points in a flat plane is a straight line. By reflecting $Q$ to $Q'$, we have "unfolded" the path.

The path $P 	o R 	o Q'$ is now a single, continuous line segment. If $P, R,$ and $Q'$ are collinear, the distance is minimized. We have turned a complex minimization problem into a simple task of finding the equation of a line.

Using the two-point form of a line equation,

$$y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}(x - x_1)$$

We substitute our coordinates $P(-3, 4)$ and $Q'(0, -2)$:

$$y - (-2) = \frac{4 - (-2)}{-3 - 0}(x - 0)$$

Simplifying this, we get $y + 2 = \frac{6}{-3}x$, which reduces beautifully to:

$$y = -2x - 2$$

### Phase 3: Locating the Optimal Point

Now that we have the equation of our path, finding point $R$ is trivial. Point $R$ lies on the $x$-axis, which means its $y$-coordinate is $0$.

Setting $y = 0$ in our equation $y = -2x - 2$, we get $0 = -2x - 2$, which leads us directly to $x = -1$. Thus, our optimal point $R$ is $(-1, 0)$.

### Phase 4: The Final Calculation

We are almost there. The problem asks for the value of $50((PR)^2 + (RQ)^2)$. Let us calculate these squares.

For $PR$, with $P(-3, 4)$ and $R(-1, 0)$:

$$(PR)^2 = (-1 - (-3))^2 + (0 - 4)^2 = (2)^2 + (-4)^2 = 4 + 16 = 20$$

For $RQ$, with $R(-1, 0)$ and $Q(0, 2)$:

$$(RQ)^2 = (0 - (-1))^2 + (2 - 0)^2 = (1)^2 + (2)^2 = 1 + 4 = 5$$

Finally, we compute the expression:

$$50(20 + 5) = 50(25) = 1250$$

The final answer is **1250**. You didn't just solve a problem; you mastered a technique. Keep this "Reflection Principle" in your toolkit—it will serve you well in optics, mechanics, and beyond.

Enter Numerical Value:

Visualized Solution

Visualizing the Problem

The Reflection Principle

Finding the Shortest Path

Equation of Line $P Q^{'}$

Simplifying the Line Equation

Locating Point $R$

The Optimal Path

Calculating $(P R)^{2}$

Evaluating $(P R)^{2}$

Calculating $(R Q)^{2}$

Evaluating $(R Q)^{2}$

Final Computation

The Sigma Insight: Various Forms of Equations of a Line

The Geometry of the Shortest Path

Phase 1

Phase 2

Phase 3

Phase 4

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