Let a ray of light passing through the point (3, 10) reflects on the line 2x + y = 6 and the reflect

Q: Wait, why do we reflect point B instead of point A? Does it matter?

It does not matter! You can reflect either point. If you reflect $A$ across the mirror, the image $A'$ will lie on the line containing the reflected ray. The logic remains identical: the incident ray passes through $A$ and the point of reflection, while the reflected ray passes through the point of reflection and $B$. By reflecting $B$ to $B'$, you create a straight line $AB'$ that represents the path of the light. Both methods are perfectly valid.

Q: Why can't we just use the law of reflection (angle of incidence = angle of reflection) directly?

You absolutely can, but it is a trap! Using trigonometry to equate angles involves calculating slopes, taking tangents, and solving complex equations with square roots. It is prone to algebraic errors. The reflection principle (using the image point) is a geometric shortcut that turns a trigonometry problem into a simple linear equation problem. In JEE, efficiency is key!

Q: What if the constant term in the line equation didn't match the form $ax + by + 1 = 0$?

Great observation! If your derived equation was, for example, $6x - 2y + 2 = 0$, you would need to divide the entire equation by $2$ to make the constant term $1$. Always ensure your equation is normalized to the form required by the question before comparing coefficients.

Question

Let a ray of light passing through the point $(3, 10)$ reflects on the line $2x + y = 6$ and the reflected ray passes through the point $(7, 2)$. If the equation of the incident ray is $ax + by + 1 = 0$, then $a^2 + b^2 + 3ab$ is equal to_________

Sigma Percentile · Accepted Answer

### The Geometry of Light: A Masterclass in Reflection

Imagine you are standing in a dark room, holding a laser pointer. You aim it at a mirror, and the beam bounces off, hitting a target on the wall.

In coordinate geometry, this is not just a physical phenomenon; it is a beautiful puzzle of symmetry. Today, we are going to solve this problem not by wrestling with angles, but by mastering the elegance of the reflection principle.

### Phase 1: The Geometric Insight

We are given a reflecting line (our mirror) defined by $2x + y - 6 = 0$. A light ray passes through point $A(3, 10)$, strikes the mirror, and reflects to pass through point $B(7, 2)$.

The question asks for the equation of the incident ray. If we try to use the law of reflection—where the angle of incidence equals the angle of reflection—we would be forced into a nightmare of trigonometric slopes.

Instead, let us use the **'Mirror Image' property**. If we find the image of point $B$ across the mirror line, let us call it $B'$, then the incident ray must pass through both $A$ and $B'$. This turns a complex optics problem into a simple task of finding the equation of a line passing through two points.

### Phase 2: The Mathematical Machinery

To find the image $B'(x, y)$ of point $B(7, 2)$ across the line $2x + y - 6 = 0$, we use the standard image formula:

$$\frac{x - 7}{2} = \frac{y - 2}{1} = -2 \left( \frac{2(7) + 1(2) - 6}{2^2 + 1^2} ight)$$

Take a deep breath. This formula is your best friend in coordinate geometry. It encapsulates the perpendicularity of the construction line and the equidistant property of reflection.

Let us simplify the constant ratio on the right side. The numerator is $14 + 2 - 6 = 10$, and the denominator is $4 + 1 = 5$. Thus, the ratio becomes $-2 	imes (10 / 5) = -4$. This constant $-4$ is the key that unlocks the coordinates of $B'$.

### Phase 3: Calculating the Image Point

Now, we solve for $x$ and $y$ individually:

For $x$: 
$$\frac{x - 7}{2} = -4 \Rightarrow x - 7 = -8 \Rightarrow x = -1$$

For $y$: 
$$\frac{y - 2}{1} = -4 \Rightarrow y - 2 = -4 \Rightarrow y = -2$$

We have successfully crossed the mirror! Our image point $B'$ is at $(-1, -2)$.

### Phase 4: The Final Line

Now, the incident ray is simply the line connecting $A(3, 10)$ and $B'(-1, -2)$. First, let us find the slope $m$:

$$m = \frac{10 - (-2)}{3 - (-1)} = \frac{12}{4} = 3$$

Using the point-slope form $y - y_1 = m(x - x_1)$ with point $A(3, 10)$:

$$y - 10 = 3(x - 3) \Rightarrow y - 10 = 3x - 9 \Rightarrow 3x - y + 1 = 0$$

Comparing this to the given form $ax + by + 1 = 0$, we identify $a = 3$ and $b = -1$. Finally, we calculate the requested expression:

$$a^2 + b^2 + 3ab = (3)^2 + (-1)^2 + 3(3)(-1) = 9 + 1 - 9 = 1$$

And there it is! The **final answer is 1**. By trusting the geometry, we bypassed the complexity and arrived at the solution with precision. Keep this reflection trick in your toolkit—it is a classic JEE winner.

Enter Numerical Value:

Visualized Solution

Visualizing the Reflection Setup

The Reflection Principle

Setting up the Image Formula

Raw Setup (Substitution)

Simplifying the Constant Ratio

Calculating Coordinates of $B^{'}$

Finding the Incident Ray's Slope

Forming the Line Equation

Identifying Coefficients $a$ and $b$

Final Calculation

The Sigma Insight: Various Forms of Equations of a Line

The Geometry of Light

Phase 1

Phase 2

Phase 3

Phase 4

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